Let's assume that the feet of $A$, $B$, $C$ correspond to points $D$, $E$, $F$, respectively. We apply an inversion with radius $\sqrt{AH \cdot AD}$. In this inversion, points $D$, $H$, as well as $F$, $B$, and $E$, $C$, interchange with each other. We aim to prove that point $P$ goes to the nine point center of $ABC$ ($N_9$). We use directed angles modulo $180^\circ$. Let's suppose that the image of $P$ after inversion is $P'$ and $M$, $N$ be the midpoints of the sides $AC$ and $AB$.

$$\angle CPH + \angle HAC = \angle DP'E = 4\angle CBH$$
$M$ is the center of the cyclic quadrilateral $ABDE$, so:
$$\angle DME = 2\angle HBC$$
Also $N_9$ is the circumcenter of $MDE$. Which implies:
$$\angle DN_9E = 2\angle DME = \angle DP'E$$
In the same way we obtain:
$$\angle FN_9D = \angle FP'D$$
Finally we can conclude $P' = N_9$.

Now we will find the images of $X$, $Y$ after inversion. Let $X'$, $Y'$ be these images. As $CH$ is the perpendicular bisector of $PY$, after the inversion the circumcircle of $ADE$ is the $A$-Apollonian circle in the triangle $AY'N_9$. The center of this circle is $M$ so $M$, $N_9$, $Y'$ are collinear and:
$$MA^2 = MN_9 \cdot MY'$$
$AS$ becomes the $A$-altitude in $AX'Y'$.

We claim $ABC \sim AY'X'$. First we calculate the ratio $\frac{AY'}{AX'}$.
$$AMN_9 \sim Y'MA \implies \frac{AY'}{AN_9} = \frac{AM}{MN_9}$$
Similarly
$$\frac{AN_9}{AX'} = \frac{NN_9}{AN} \implies \frac{AY'}{AX'} = \frac{AM}{AN} \cdot \frac{NN_9}{MN_9} = \frac{AB}{AC}$$
Then we calculate the angle $X'AY'$.
$$\angle X'AY' = \angle Y'AB + \angle X'AC - \angle BAC = \angle AN_9M + \angle AN_9N - \angle BAC = \angle BAC.$$
Hence the claimed similarity is proven. Now let the line perpendicular to $X'Y'$ be $\ell$. We use again directed angles modulo $180^\circ$.
$$\begin{aligned} \angle \ell, AC &= \angle AX', AC = \angle ACB - 90^\circ + \angle NN_9A \\ &= \angle ACB + \angle NN_9, AC + \angle AC, AN_9 - 90^\circ = \angle BAC + \angle CAN = \angle BAN_9 \end{aligned}$$
The final equality proves the desired result. ■