Actually, it works for all kinds of triangles (is not necessary to have $\Delta$ ABC a right one)

I have seen this issue before, so you can search. Reflect C on AB, name it C'. Any circle inscribed in the circular segment CBB' has A Q R collinar, next see $ACQ \sim ARC$ so $AC^2=AQ*AR=AP^2$

MY HINT: lemma:let $W$ be a circle and let $W1$ be a circle in $W1$ such that $W1$ touches $W$ in $S$.let segment $AB$ such that $AB$ is the chord of $W$ and touches $W1$ in $Q$.let $R$ be the intersection of $PQ$ and $W$.then we have: $\angle{ARS}=\angle{SRB}$ proof:it is obvious! HINT:by the lemma prove that $A$,$Q$ and $R$ are collinear and for $AP=AC$ prove that $AC^2=AQ.AR$ or $\angle{ACF}=\angle{ARC}$ then we are done!

Here's a really short way to prove that $A$, $Q$, and $R$ are collinear. I have no idea if this method can be used to prove $AP=AC$, but here goes. Let $\ell$ be the tangent to $(ABC)$ at $A$. Note that $AB\perp \ell$, so $CF\parallel \ell$. Now remark that because $\omega$ and $(ABC)$ are internally tangent, there exists a homothety taking the former to the latter, and due to the established parallelism it also takes $Q$ to $A$. Hence $A$, $Q$, and $R$ are collinear as desired.

Let $\omega$ be the circumcircle of $\triangle{ABC}$ and $RP \cap \omega = M$ It is easy to see $AM=BM$ Hence $\angle{ARM}=\frac{\pi}{4}$ $\angle{QFP}=\frac{\pi}{2} \implies \angle{QRP}=\frac{\pi}{4}$ Hence $A,Q,R$ are collinear. $\angle{QRB}=\angle{QFB}=\frac{\pi}{2}$ $\implies Q,R,B,F$ are concyclic $\implies AQ.AR=AF.AB$ $\implies AP^2=AC^2$ $\implies AP=AC.$

sasanineq solution in latex form and a few changings: lemma:let $W$ be a circle and let $W1$ be a circle in $W1$ such that $W1$ touches $W$ in $S$.let segment $AB$ such that $AB$ is the chord of $W$ and touches $W1$ in $Q$.let $R$ be the intersection of $PQ$ and $W$.then we have: angle $ARS$= $SRB$ proof:it is obvious! by the lemma $A$, $Q$, $R$ are collinear because $A$ is the midpoint of arc $CQ$ ($Q$ is the intersection of $FC$ with circumcircle of $ABC$) and for proving $AP=AC$ prove that $AC^2$=$AQ.AR$ and $AP^2=AQ*AR$ and $FQRB$ is cyclic.

also here