Let the circles with diameter $AC$ and $BD$ meet at $M$ and $N.$ By Steiner, it's enough need to show that $MN$ passes through $X$ iff it passes through $Y.$ Invert with pole $Q$ and arbitrary radius. Suppose that after inversion point $W$ goes to $W'.$ Circles with diameter $AC$ and $BD$ go to circles with diameter $A'C'$ and $B'D',$ respectively. Also, $X'$ and $Y'$ go to the intersections of $A'B'$ with $C'D'$ and $B'C'$ with $A'D'.$ By Steiner, circles with diamter $X'Y', A'C'$ and $B'D'$ are coaxial, hence $X'Y'M'N'$ are concyclic, thus $X'$ lies on $QM'N'$ iff $Y'$ lies on $QM'N'.$

First of all, notice that $X$ is the Miquel Point of complete quadrilateral $ABDCPQ$, so it lies on circles $(PAC)$ and $(PBD)$. The key result we use to solve this problem is that $H_1H_2$ is the Aubert line, so it is perpendicular to the Newton-Gauss line. Hence, if we let $I$ and $J$ be the midpoints of $AC$ and $BD$ respectively, the condition $X \in H_1H_2$ rewrites as : $XI^2 - XJ^2 = H_1I^2 - H_1J^2$ By orthogonality ($H_1$ is the orthocenter of $\triangle PAD$) and the Median's Theorem, this becomes : $XI^2 - XJ^2 = (AC^2 - BD^2)/4$ By the Median's Theorem again and Cosine Law, this rewrites as : $2 \cdot XA \cdot XC \cdot cos \angle AXC = 2 \cdot XB \cdot XD \cdot cos \angle BXD$ As triangles $AXC$ and $BXD$ are similar (spiral similarity), the previous relation is equivalent to the last $2$ triangles being congruent, which is in turn equivalent to $AC = BD$, because $\angle AXC = \angle BXD$ are obtuse, by hypothesis ($\angle AXC = \pi - \angle APC$ via the cyclicity of $APCX$). In conlusion, $X \in H_1H_2$ and $AC = BD$ are equivalent. Analogously, $Y \in H_1H_2$ and $AC = BD$ are equivalent. Hence if $X$ lies on $H_1H_2$, then $Y$ also lies on $H_1H_2$, which is exactly what we wanted to prove.

Attachments:

Hilarious problem. Relabel point $X$ as point $T$. Let $\triangle EFG$ be the $\triangle PBC$ orthic triangle and $\triangle XYZ$ be the $\triangle PAD$ orthic triangle. Let $\omega_1$ and $\omega_2$ be the circles with diameters $AC$ and $BD$, with centers $M$ and $N$ the midpoints of $AC$ and $BD$. By cyclic quadrilaterals $BGFC$ and $AZYD$ from the orthocenters, Power of a Point tells us that\[GH_1 \cdot CH_1 = BH_1 \cdot FH_1 \text{ and } ZH_2 \cdot DH_2 = YH_2 \cdot AH_2\]hence $H_1H_2$ is the radical axis of $\omega_1$ and $\omega_2$. Note that $T$ is a center of spiral similarity sending $AC \to BD$ and so is $Y$. Recall that\[T \in H_1H_2 \implies \text{Pow}_{\omega_1} T = \text{Pow}_{\omega_2} T \implies TM^2 - MC^2 = TN^2 - ND^2\]but since $\triangle TNC \sim \triangle TND$ by the spiral similarity, it follows that this can only be true if the spiral similarity is a congruency and $AC = BD$. Similarly, recall that\[Y \in H_1H_2 \implies \text{Pow}_{\omega_1} Y = \text{Pow}_{\omega_2} Y \implies YM^2 - MC^2 = YN^2 - ND^2\]but since $\triangle YNC \sim \triangle YND$ by the spiral similarity, it follows that this can only be true if the spiral similarity is a congruency and $AC = BD$ as well. Hence $T \in H_1H_2 \iff Y \in H_1H_2$, which is actually a bit stronger than what we wanted to prove. $\blacksquare$