On equivalence of P-Q being a projection with the other statements. I am trying to do the following exercise from Conway's FA

If Ρ and Q are projections, then the following statements are equivalent, (a) P — Q is a projection, (b) ran Q \subset ran P. (c) ΡQ = Q. (d) QΡ = Q.

If P Q = Q then because a projection is hermitian then QΡ = Q can be achieved by simple calculation.
Also equivalence of each of ΡQ = Q and QΡ = Q with P-Q being a projection is a matter of calculations considering projection=idempotent+hermitian.
Also, The question $q-p$ is a projection when $pq = p$  only addresses partially only one implication: d --> a.
How can the other parts be done?
 A: Recall that, as in  Conway's "Functional Analysis" book:

Definition. An idempotent operator on Hilbert space $H$ is a bounded linear operator $E$ on $H$
such that $E^2 = E$. A projection is an idempotent operator $P$ such that $\ker P = ( \operatorname{ran} P)^{\perp}$.

We will start with two lemmas:
Lemma 1 If $P$ is  idempotent, then $P$ is a projection if and only if, for all $x, y \in H$, $\langle x- Px, Py \rangle=0$.
Proof: It is immediate. Just note that $\operatorname{ker} P =\{x -Px : x \in H\}$ and that $H= \operatorname{ker} P  \oplus \operatorname{ran} P$. $\square$
Lemma 2 If $P$ is  idempotent, then $P$ is a projection if and only if, $P$ is self-adjoint.
Proof:
$(\Rightarrow)$  Suppose $P$ is a projection. Then, by lemma 1, we have that
for all $x, y \in H$, $\langle x- Px, Py \rangle=0$. So, for all $x, y \in H$,
$$ \langle x, (1-P)^*Py \rangle=\langle x- Px, Py \rangle=0 $$
So, for all $ y \in H$, $(1-P)^*Py=0$. So, $(1-P)^*P=0$. So, we have
$$ P-P^*P= (1-P^*) P = (1-P)^*P=0$$
So, $P= P^*P$.  But, then $P^* =(P^*P)^*= P^*(P^*)^*= P^*P = P$. So $P$ is self-adjoint.
$(\Leftarrow)$ Suppose $P$ is self-adjoint. Then, for all $x, y \in H$, $$\langle x- Px, Py \rangle= \langle P^*x- P^*Px, y \rangle= \langle Px- P^2x, y \rangle= \langle 0, y \rangle =0$$
So, by lemma 1, $P$ is projection. $\square$
Now let us prove that:

If $P$ and $Q$ are projections, then the following statements are equivalent: (a) $P-Q$ is a projection. (b) $ \operatorname{ran}Q \subseteq  \operatorname{ran}P$. (c) $PQ=Q$. (d) $QP=Q$.

Proof: (b $\Leftrightarrow$ c).
Suppose $ \operatorname{ran}Q \subseteq  \operatorname{ran}P$. Then, for all $x \in H$, $Qx \in  \operatorname{ran}Q \subseteq  \operatorname{ran}P$. So, there is $y \in H$ such that $Qx =Py$. So $ PQx=P^2y=Py=Qx$. So $PQ= Q$.
Suppose $PQ=Q$. Then, if $y\in  \operatorname{ran}Q$, there is $x\in H$ such that $ y= Qx = PQx$. So $y \in  \operatorname{ran}P$. So, $ \operatorname{ran}Q \subseteq  \operatorname{ran}P$.
(c $\Leftrightarrow$ d)
Suppose $PQ=Q$. Since, by lemma 2, $P$ and $Q$ are self-adjoint, we have
$$QP= Q^*P^* = (PQ)^* = Q^* = Q$$
Suppose $QP=Q$.  Since, by lemma 2, $P$ and $Q$ are self-adjoint, we have
$$PQ = P^*Q^* = (QP)^*= Q^* = Q$$
(d $\Leftrightarrow$ a)
(d $\Rightarrow$ a) Suppose $QP=Q$. Then, by (c $\Leftrightarrow$ d), we also have that  $PQ=Q$. So, we have
\begin{align*}
(P-Q)(P-Q) & = P^2 - PQ - QP + Q^2 =\\
&= P -Q -Q +Q = \\ 
&= P-Q
\end{align*}
So $P-Q$ is idempotent. Since, by lemma 2, $P$ and $Q$ are self-adjoint, we have that $P-Q$ is self-adjoint. So, by lemma 2, $P-Q$ is a projection.
(a $\Rightarrow$ d). Suppose that $P-Q$ is a projection. Then $P-Q$ is idempotent, so
\begin{align*}
P-Q &= (P-Q)(P-Q) = \\
& = P^2 - PQ - QP + Q^2 = \\
&= P -PQ -QP +Q 
\end{align*}
So,
$$2Q = PQ+QP $$
Since $Q$ is idempotent, we have
$$2Q = 2Q^2= QPQ + Q^2P= QPQ +QP$$ and   $$2Q = 2Q^2= PQ^2 + QPQ= PQ +QPQ$$ It follows that $ QPQ +QP = PQ +QPQ$. So  $QP=PQ$.
So we have $2Q =  PQ+QP = 2QP$. It means $QP = Q$. $\square$
