Is Projective Geometric Algebra a strict subset of Conformal Geometric Algebra? My understanding is that CGA has an additional basis vector $e_{\infty}$, so it would seem to me that if that is the only difference then PGA fits entirely within CGA.
However I don't know if the introduction of the new basis affects some of the operations, since now the pesudoscalar of PGA is not a pseudoscalar in CGA, since there are 5 grade 4 blades.
 A: $
\newcommand\Cl{\mathrm{Cl}}
\newcommand\R{\mathbb R}
\newcommand\flat\mathbf
\newcommand\origin{{\oslash}}
\newcommand\lcontr{\mathbin\rfloor}
\newcommand\rcontr{\mathbin\lfloor}
\newcommand\flatb\boldsymbol
\newcommand\rev\widetilde
\newcommand\cliff\overline
\newcommand\dual{\mathop\star}
\newcommand\grd\widehat
\newcommand\revdual{\mathop{\rev\star}}
\newcommand\cliffdual{\mathop{\cliff\star}}
\newcommand\mdual{\mathop*}
\newcommand\invmdual{\mathop{*^{-1}}}
\newcommand\infdot{\mathop{\underset\infty\cdot}}
$Yes. It is shown how this the case both for PGA and dual PGA in Projective Geometric Algebra as a Subalgebra of Conformal Geometric algebra by Ales Navrat, Jaroslav Hrdina, Petr Vasik, and Leo Dorst. However, this paper seems to me to be very scant on geometry, relying instead on demonstrating the formal algebraic relationship. I would like to bring geometry back into this.
PGA
CGA takes place in the algebra $\Cl_{4,1}(\R)$; let the space of vectors be denoted by $V$. We choose $\infty \in V$ with $\infty^2 = 0$ and declare it to be the point at infinty. We then define a point to be (a multiple of) $\infty$ or any $x \in V$ such that $x^2 = 0$ and $x\cdot\infty \ne 0$. Two vectors represent the same point if they are multiples of each other; a point is normalized if $x\cdot\infty = 1$ or $x = \infty$. From here on we assume all points are normalized unless otherwise stated. A fat point is just like a point except $x^2 \ne 0$; we call $2x^2$ the squared radius of the fat point when $x^2 > 0$. Notice that we can change the radius of a point by adding a multiple of $\infty$:
$$
  \left(x + \frac12a\infty\right)^2 = a.
$$
If $I$ is a unit pseudoscalar then $xI$ is a sphere when $x$ is a fat point.
If $x, y, z \in V$ are points then recall that
$$
  x\wedge y\wedge\infty,\quad x\wedge y\wedge z\wedge\infty
$$
are respectively the line through $x, y$ and the plane through $x, y, z$ (constructed by considering them as generalized spheres that pass through $\infty$). Flats are always constructed by intersecting $\infty$, so by analogy we consider
$$
  x\wedge\infty
$$
as a candidate for flat points. Indeed, this makes much sense: because
$$
  (x + a\infty)\wedge\infty = x\wedge\infty
$$
when $a$ is a scalar these flat points are insensitive to the radius of the point $x$. In particular this means that the difference of two flat points
$$
  (x - y)\wedge\infty
$$
becomes a good model for displacements.
It's in this way that we have already discovered PGA. Notice that a $k$-flat is represented by a $(k+1)$-vector; hence the dual of a $k$-flat has grade $5 - (k+1) = 4-k$ which is exactly the PGA grade of a flat. The vectors $\flat x \in V$ which dualize to planes must pass through $\infty$, so
$$
  0 = \infty\wedge(\flat xI) = (\infty\cdot\flat x)I \iff \infty\cdot\flat x = 0.
$$
The orthogonal complement of $\infty$ is necessarily the span of $\infty$ and a $3$-space with signature $(3,0,0)$, which all together is a $4$-space with signature $(3,0,1)$. This is exactly the signature of PGA.
From here on, I will use the convention that bold letters $\flat X$ represent PGA objects, while unbolded letters $X$ represent "CGA objects" (with the special exception of $\infty$ being both). So if I say "a plane $\flat X$" then $\flat X$ is a vector, but if I say "a plane $X$" then $X$ is a $4$-vector, and $\infty$ will be referred to both as a plane and as a point.
A CGA pseudoscalar $I$ actually immediately gives rise to a PGA pseudoscalar: notice that $\infty I^{-1}$ is a $4$-vector and that
$$
  \infty\lcontr(\infty I^{-1}) = (\infty\wedge\infty)I^{-1} = 0
$$
so $\infty I^{-1}$ is orthogonal to $\infty$ and hence a $4$-vector in the PGA space, and hence a PGA pseudoscalar; denote this by $\flat I = \infty I^{-1} = -\infty I$. If we choose any CGA point $\origin$, then we can invert this procedure and get $I$ from $\flat I$:
$$
  -\origin\wedge\flat I = \origin\wedge(\infty I) = (\origin\cdot\infty)I = I.
$$
This embedding of PGA in CGA gives an effective way of dealing with CGA flats $X$: translate them to a PGA flats $XI$, perform PGA operations to get some $\flat Y$, and then transform back via $\flat YI^{-1}$. The case of points $x$ is slightly exceptional since our standard representation of points in CGA isn't flat. In this case we transform into PGA via $(x\wedge\infty)I$ and transform a PGA point $\flat Y$ back via
$$
  |\flat Y|(\flat YI^{-1})\rcontr\origin - \frac12\Bigl[(\flat YI^{-1})\rcontr\origin\Bigr]^2\infty
  = |\flat Y|(\flat Y\wedge\origin)I^{-1} + \frac12\Bigl[\flat Y\wedge\origin\Bigr]^2\infty.
\tag{Pt}
$$
with $|\flat Y| = \sqrt{-Y^2}$ and $\origin \ne \infty$ an arbitrary normalized CGA point. The expression $(\flat Y\wedge\origin)I^{-1}$ will become quite interesting down below. Notice that we lose information here: all ideal PGA points get mapped to the CGA point $\infty$, so it is best to stick with flat points as much as possible.
PGA$^*$
We now turn our attention to dual PGA, which I will denote PGA$^*$. Just like PGA models Euclidean geometry, dual PGA models dual Euclidean geometry where all "statements" of Euclidean geometry remain the same but we swap the words "point" and "plane". This is simplest to describe in the 2D case, where points are dual to lines: in Euclidean geometry, points have a distance between them, lines have angles between them, and lines can be perpendicular to each other; in dual Euclidean geometry, lines have distance between them, points have angles between them, and points can be perpendicular to each other. Instead of a plane at infinity, we have a point at infinity. PGA$^*$ is an algebra isomorphic to PGA, but we interpret vectors as points instead of planes. It should be intuitive that the relationship between PGA$^*$ and PGA is related to projective duality. We essentially get PGA$^*$ be applying projective duality and then swapping the words "point" and "plane"; for instance
$$\begin{aligned}
  &\quad
  \text{There is a unique distance between two points}
\\
  \overset{\text{dual}}\longmapsto &\quad
  \text{There is a unique angle between two planes}
\\
  \overset{\text{swap}}\longmapsto &\quad
  \text{There is a unique angle between two points}
\end{aligned}$$
The reason we need dual and swap (and not just swap) is to get the correct grade; in the above, we take trivectors to vectors to vectors.
We can realize projective duality as follows: fix an origin point $\flatb\origin$ in PGA. We wish to take a point $\flat x$ with distance $d = |{\flatb\origin}\vee{\flat x}|$ from $\flatb\origin$ to a plane orthogonal to the line ${\flatb\origin}\vee{\flat x}$ a distance $1/d$ away from, and vice versa. The correct map for any flat ${\flat x}$ turns out to be
$$
  \dual{\flat x} = {\flat x}{\flat I} + ({\flatb\origin}\vee{\flat x}){\flatb\origin}^{-1}
$$
where we assume $\flat I = \flatb\origin\infty$. We could replace $+$ with $-$, where one choice places $\dual\flat x$ on the "same side" of $\flatb\origin$ as $\flat x$, the other on the opposite side; I'm not going to confidently state which is which. This map is an involution:
$$
  \dual(\dual{\flat x}) = {\flat x}.
$$
It is intimately related to the Hodge star: the operations
$$\begin{aligned}
  \dual({\flatb\origin}\vee\cliff{\flat y}),\quad
  \cliff{(\dual{\flat y})\wedge\infty},\quad
  \flatb\origin\vee(\dual\rev{\flat z}),\quad
  \revdual({\flat z}\wedge\infty)
\end{aligned}$$
are exactly the Hodge star and (respectively) its inverse on ideal elements $\flat y$ and flats $\flat z$ intersecting $\flatb\origin$. Here, $\cliff{\flat y}$ is Clifford conjugation (reversal and grade involution), $\rev{\flat y}$ is reversal, and $\revdual$ is reversal applied to the result of $\dual$ (since I don't know how to get a really wide tilde with MathJax).
I haven't clearly defined the join product $\vee$. Doing so requires delicacy I don't want to employ at the moment, so I will side step the issue by noting that
$$
  \flatb\origin\vee\flat x = \flat x\rcontr\origin
$$
where $\origin$ is the CGA point corresponding to $\flatb\origin$. We can then do some symbol pushing and find the following CGA expression for $\dual\flat x$:
$$
  \dual\flat x = -(\origin + \infty)^{-1}(\flat x\wedge\origin)I^{-1}(\origin + \infty).
\tag{CGAPD}
$$
Notice that the $(\flat x\wedge\origin)I^{-1}$ term is the same one that is in (Pt). The above equation is wonderfully suggestive. The transformation
$$
  x \mapsto \sigma(x) = -(\origin + \infty)^{-1}x(\origin + \infty)
$$
is a homomorphism which swaps $\origin$ and $\infty$ and leaves their orthogonal complement fixed. We can understand this as swapping the plane $\infty$ with the point $\origin$, and so this represents the "swap" step in dualizing PGA. If this is the swap step and $\dual$ is projective duality, then it immediately follows that
$$
  \mdual\flat x = \sigma(\dual\flat x) = (\flat x\wedge\origin)I^{-1}
$$
is the geometry-preserving dualization between PGA and PGA$^*$ (i.e. maps points to points, planes to planes, etc.). An easy calculation shows that its inverse is
$$
  \invmdual x = (x\wedge\infty)I.
$$
Recall that a CGA point $x$ can be written
$$
  x = \origin + x_3 - \frac12x_3^2\infty
$$
where $x_3$ is a vector in the Euclidean space orthogonal to $\origin$ and $\infty$. It follows that the corresponding PGA$^*$ point is
$$
  x + \frac12x_3^2\infty
$$
which is the fat point centered at $x$ with squared radius $x_3^2$. This gives PGA$^*$ the curious interpretation as an algebra of sphere intersections where the spheres always have radius equal to the distance their center is from the origin; for instance, the line between two PGA$^*$ points can be interpreted as the circle of intersection of such spheres. What's happening is that the CGA interpretation of $\sigma$ is as an involution through $\origin$; the planes of PGA involute into into spheres through the origin and are then interpreted as a representation their center points.

Strictly speaking, we do not need to apply $\sigma$; we could take the perspective that PGA$^*$ is the same algebra as PGA but reinterpreted via $\dual$. It's really the form of (CGAPD) that compels me to consider $\sigma\circ\dual$, as well as the fact that this map respects CGA's geometric primitives: the map $x \mapsto x\wedge\infty$ on this view of PGA$^*$ takes PGA$^*$ flats immediately to the corresponding CGA flats.
However, the "same algebra" viewpoint is very fruitful, especially if we want to consider PGA independently. In particular, this allows us to define the join product $\vee$. Though this may seem to be circular since my definition of $\dual$ involves $\vee$, we can reformulate the definition to avoid its use as follows; this reformulation will also make the connection with the Hodge star exceedingly clear.
Fix a point $\flatb\origin$ in PGA. We notice that for any ideal $\flat x$ there is a unique $\flat x'$ containing $\flatb\origin$ (i.e. $\flat x'$ is in the subalgebra with $\flatb\origin$ as "pseudoscalar") such that
$$
  \flat x'\infty = \flat x.
$$
This allows us to define a scalar product $\infdot$ on ideal elements:
$$
  \flat x\infdot\flat y = \flat x'\cdot\flat y'.
$$
You can confirm that in this case $\infdot$ is independent of the point $\flatb\origin$ chosen. We extend this $\infdot$ to the whole algebra by noting that an arbitary $\flat x$ splits as
$$
  \flat x = \flat x_\origin + \flat x_\infty,\quad \flat x_\origin = \flat x\lcontr\flatb\origin\,\flatb\origin^{-1}
$$
where $\flat x_\origin$ contains $\flatb\origin$ and $\flat x_\infty$ is ideal. Then
$$
  \flat x\infdot\flat y = \flat x_\infty\infdot\flat y_\infty.
$$
This is not independent of the choice of $\flatb\origin$. From here, $\infdot$ extends to the whole algebra by linearity.
Now we may define $\dual\flat x$ for any $\flat x$ much like we would define the Hodge star: it is the unique element such that for all $\flat y$
$$
  \flat y\wedge(\dual\flat x) = \flat y\cdot\flat x\,\flat I + \flat y\infdot\flat x\,\flat I'
$$
where $\flat I = \flatb\origin\infty$ and $\flat I' = \flatb\origin^{-1}\infty$.
