Straightforward length bash work. Let $AB = 2a$, then $AM = MC = a$. Note that $P$ is just the projection of $A$ wrt $BM$. and we could easily get $BP = \frac{4a}{\sqrt{5}}$ and $PM = \frac{a}{\sqrt{5}}$. Then, let $BP \cap (APC) = K$. We could get by POP that \[ PM \cdot MK = AM \cdot MC \implies MK = a \sqrt{5} \]Therefore, just note that \[ BP \cdot BK = \frac{4a}{\sqrt{5}} \cdot 2a \sqrt{5} = 8a^2 = (2a \sqrt{2})^2 = BC^2 \] Here's a synthetic solution. Reflect $B$ wrt $M$ to get $K$. Then, note that $\angle APB = 90^{\circ}$, so $\angle APK = 90^{\circ}$. But we know that $ABCK$ is a parallelogram since $MA = MC$ and $MB = MK$. Therefore, $\angle ACK = 90^{\circ}$. So, $APCK$ cyclic. To finish, just note that \[ BP \cdot BK = BP \cdot 2BM = 2 \cdot BA^2 = BC^2 \]Note that $BP \cdot BM = BA^2$ is pop on $(APM)$ from point $B$.

Consider $O$ the midpoint of $BC$ clearly $ABOP$ it's cyclic because $\angle BOA = 90^{\circ}$ = $\angle BPA$ , how $ABC$ it's isosceles $AO$ it's also the angle bissector of $\angle BAC$ , then $\angle BAO = 45^{\circ}$ = $\angle BPO$, observe that $\angle MPO = 135^{\circ}$ and $\angle MCO = 45^{\circ}$ implies $OPMC$ cyclic. Finally just note that $\angle OMC = 90^{\circ}$ = $\angle OPC$ give that $\angle PCO$ = $\angle POA$ = $\angle PBA$ and clearly $\angle PBA$ = $\angle PAC$ ($APM$ and $ABM$ are similar) then $\angle PAC$ = $\angle PCO$.

WLOG $AC=2$, let $A(0,0)\Rightarrow B(0,2), C(2,0)$, let $M(1,0)$ the midpoint of $AC$ and $P=(x,y)$, the equation for the circle $\Gamma$ with center $(0,1)$ and radius 1 is $$(x-0)^2+(y-1)^2=1^2\Rightarrow x^2+(y-1)^2=1$$ And the equation of $BM$ is $$y=\frac{2-0}{0-1}x+0\Rightarrow y=-2x+2$$ Solving the sistem we have $x=\frac{4}{5}, y=\frac{2}{5}\Rightarrow P\left(\frac{4}{5}, \frac{2}{5}\right)$ If $O$ is the center of $(APB)$, $O$ is of the form $O(1,k)$ (since $O$ is on the perpendicular bisector of AB) Then $$AO=PO\Leftrightarrow (0-1)^2+(0-k)^2=(\frac{4}{5}-1)^2+(\frac{2}{5}-k)^2\Leftrightarrow k=1$$ Finally, is enough to prove $BC^2=BO^2-r^2$ where $r$ is the radios of $(APB)$, thus $$BC^2=BO^2-r^2\Leftrightarrow (0-2)^2+(2-0)^2=(0-1)^2+(2-(-1))^2-\left[(0-1)^2+(0-(-1))^2\right] \Leftrightarrow 8=8$$ And we are done

Notice that $AP'CP$ is a deltoid and not a parallelogram, and AP is perpendicular to $\ell$. jampm wrote:

Let $A=(0,0)$, $B=(0,2)$, $C=(2,0)$ in the Cartesian plane. We easily obtain $\ell:y=-2x+2$ and $\Gamma:x^2+(y-1)^2=1$. Solving this system, we find $P=\left(\frac45,\frac35\right)$. We can verify that $(x-1)^2+(y+1)^2=2$ is the circumcircle of $ACP$, and $BC$ is $x+y=2$. It suffices to show that the only solution to the system: \begin{align*}(x-1)^2+(y+1)^2&=2\\x+y&=2\end{align*}is $(x,y)=(2,0)$. Substituting $y=2-x$, we have $(x-1)^2+(x-3)^2=2$, and so $2(x-2)^2=0$. This finally proves the result.

Let $M$ be the midpoint of $AC$, and let $D$ be the point such that $ABCD$ is a parallelogram, whose diagonals are clearly $\ell=BD$ and $AC$. Since $AB$ is diameter in $\Gamma$, we have $\angle APB=90^\circ=\angle APD$, or $P$ is on circle $\Omega$ with diameter $AD$. This circle clearly passes through $C$ because $ACD$ is by construction isoscelest at $C$ and with $\angle ACD=90^\circ$. By symmetry in the isosceles right-angled triangle $ACD$, $\Omega$ is tangent to the parallel to $AD$ through $C$, which is $BC$, and we are done.

Why so many coord bash when you can angle chase? It is sufficient to show $\angle PAC = \angle BCP$. Let $K$ denote the midpoint of $BC$. Notice that $\angle CMP = 90^{\circ} + \angle PAC$ and $\angle KMC = 90^{\circ}$. So, it is sufficient to show $\angle PMK = \angle PCK \Leftrightarrow$ $CMPK$ cyclic. First, notice that $K$ lies on $(BPA)$ since $\angle BKA = 90^{\circ}$. Now we have \[ 45^{\circ} = \angle KCA = \angle BAK = \angle PAK\]as desired. $\blacksquare$

Let $\angle PAC=\alpha ; \angle PCB=\beta$ Note that it is sufficient to show that: $\angle PAC=\angle PCB \iff \alpha=\beta$ $\implies \angle PAC=\angle PBA=\alpha \implies MP.MB=MA^2=MC^2 \implies 45-\alpha=\angle PBC=\angle PCA=45-\beta$ $\implies \alpha=\beta$.$\blacksquare$

Here is one boring approach and one overkill approach. Boring approach Set $\theta = \angle ABP = \arctan(1/2)$, so $\angle PAB = 90^{\circ}-\theta$ and $\angle CAP = \theta$. We wish to show $\angle BCP = \theta$ as well. By trigonometric Ceva, it is enough to verify \[ \frac{\sin(45^{\circ}-\theta)\sin(90^{\circ}-\theta)\sin(45^{\circ}-\theta)} {\sin\theta\sin\theta\sin\theta} = 1. \]Compute \begin{align*} \sin\theta &= \frac{1}{\sqrt5} \\ \sin(90^{\circ}-\theta) &= \cos\theta = \frac{2}{\sqrt5} \\ \sin(45^{\circ}-\theta) &= \sin(45^{\circ})\cos\theta - \cos(45^{\circ})\sin\theta = \frac{1}{\sqrt2} \cdot \frac{2-1}{\sqrt5} = \frac{1}{\sqrt{10}} \end{align*}So the numerator and denominator both equal $5^{-3/2}$ as needed. Overkill approach Define $Z$ as the second Brokard point --- in a general triangle, this point has barycentric coordinates \[ Z = \left( \frac{ab}{c} : \frac{bc}{a} : \frac{ca}{b} \right) \]thus always lying inside $ABC$, and satisfies the angle condition \[ \angle ZBA = \angle ZAC = \angle ZCB. \]Now specialize to our right triangle to conclude that: From $a=\sqrt2$, $b=c=1$ we learn $Z = (2:1:2)$, so $Z$ lies on the $B$-median. From $\angle ZBA = \angle ZAC$ follows $\angle BZA = \angle BAC = 90^{\circ}$. Hence $Z = P$. Ergo, $\angle PAC = \angle PCB$ as needed.