I would be interested to know whether this notion has been studied in this setting before.
Projective Boolean algebras have been studied in many papers, starting with Halmos's 1961 paper. I do not know whether anyone has tried to study weakly projective Boolean algebras. I tried to find something about this today on MathSciNet but I was unsuccessful. Nevertheless:
Theorem.
There exist weakly projective Boolean algebras that are not projective.
To show this, I will describe a way to generate some
weakly projective Boolean algebras, and then show
that some of them are not projective.
Lemma 1. Let $P$ be a projective algebra in some variety $\mathcal V$.
If $W$ is an algebra in $\mathcal V$ for which there exists an injective homomorphism
$\alpha\colon W\to P$ and a surjective homomorphism $\beta\colon W\to P$,
then $W$ is weakly projective.
Proof. If $A\in {\mathcal V}$
and $h\colon A\to W$ is surjective, then $\beta\circ h\colon A\to P$
is surjective. Since $P$ is projective in $\mathcal V$,
the homomorphism $\beta\circ h$ has a section, say $\sigma\colon P\to A$.
Sections of surjective homomorphisms are injective,
so the composite map $\sigma\circ\alpha\colon W\to A$
is injective. This embeds $W$ into $A$,
verifying the weak projectivity of $W$. \\\
Corollary 2. If $F$ is a free Boolean algebra of
infinite rank and $W\leq F\times F$ is the graph of a congruence
on $F$, then $W$ is a weakly projective Boolean algebra.
Proof.
If $F$ is a free Boolean algebra of infinite rank, then $F\cong F\times F$.
Since $W$ is the graph of a congruence on $F$,
$W\subseteq (F\times F)\cong F$, and the composition of the implied
maps yields an embedding $\alpha\colon W\to F$.
On the other hand, $W$ is a subdirect subalgebra of $F\times F$
since it is the graph of a congruence,
so the first coordinate projection is a surjective
homomorphism $\beta\colon W\to F$. Now apply Lemma 1 to conclude
that $W$ is weakly projective. \\\
Now the hard part. It is known that any
countable Boolean algebra of more than one
element is projective. Therefore, to exhibit
a weakly projective Boolean algebra that is not projective I must work
with uncountable Boolean algebras. For this part
I will argue that if $F$
is a free Boolean algebra of uncountable rank
and $\theta$ is a congruence on $F$ of index two,
then the graph $W$ of $\theta$ is not a projective algebra.
(By Corollary 2, this $W$ will be weakly projective.)
Lemma 3. Let $F$ be a free Boolean algebra of
uncountable rank $\kappa$
and let $\chi\colon F\to 2$ be a character/homomorphism.
Let $I=\chi^{-1}(0)$ be the prime ideal
of $F$ associated to $\chi$
and let $U=\chi^{-1}(1)$ be the ultrafilter
of $F$ associated
to $\chi$. The graph of $\ker(\chi)$ (which is
$W=(I\times I)\cup (U\times U)\leq (F\times F)$)
is not a projective Boolean algebra.
Proof.
[It is a fact that any prime ideal or ultrafilter
of a
free Boolean algebra of rank $\kappa$
requires $\kappa$ generators. In particular,
neither $I$ nor $U$ is principal.]
To prove that $W$ is not projective I will refer
to the Bockstein Separation Property (BSP), which is a property
that is satisfied by any projective Boolean algebra.
The BSP asserts that if $J$ and $K$ are ideals satisfying
$J\cap K = \{0\}$, then there exists countably generated
ideals $J'\supseteq J, K'\supseteq K$ such that
$J'\cap K'=\{0\}$. We will see that this property
fails in $W$.
Let $J = I\times \{0\}$ and let $K=\{0\}\times I$.
$J$ and $K$ are disjoint ideals of $W$.
If $W$ were projective, then by the BSP there must be
disjoint, countably generated ideals
$J'\supseteq J, K'\supseteq K$. We will complete the
argument by showing that (i) the disjointness
of $J'$ and $K'$ forces $J'=J, K'=K$, while
(ii) neither $J$ not $K$ is countably generated.
Together, (i) and (ii) refute the BSP for $W$,
so $W$ is not projective.
For Goal (i), suppose that $J'$ properly contains $J=I\times \{0\}$.
If $(a,b)\in J'-J$, then since $(a,b)\in W=(I\times I)\cup (U\times U)$,
we must have $a, b\in I$ or $a, b\in U$.
In either case, $b\neq 0$. (If $b\in U$, then $b\neq 0$ since $0\notin U$;
if $b\in I$, then $a\in I$ and
since $(a,b)\notin J$ we have $b\neq 0$.)
In either case we may choose $c<b$ such that $a, b, c\in I$
or $a, b, c\in U$. I am using that $U$ is not principal
in the latter case.
Thus, $(a,c) < (a,b)\in J'$, so $(a, c)\in J'$.
We have $(a,c) \vee (0,b - c) = (a,b)\in J'$,
$(0,b-c)\in J'\cap K \subseteq J'\cap K' = \{(0,0)\}$,
which contradicts $c<b$. This contradiction
establishes Goal (i).
For Goal (ii), assume that $J'=J=I\times \{0\}$
is a countably generated ideal
in $W$. Apply the first projection map
to this set of pairs we obtain that $I$ is countably
generated in $F$. But, as noted in the first paragraph,
$I$ requires $\kappa$ generators, and $\kappa$ is uncountable.
This completes the proof of Goal (ii).
\\\
