IMO Shortlist

Let $M$ and $N$ be the midpoints of $AD$ and $BC$, respectively and let the perpendicular bisector of $AB$ intersect the line through $P$ parallel to $AB$ at $R$.

Lemma. Triangles $QAB$ and $RNM$ are similar.

Proof. Let $O$ be the circumcentre of triangle $ABC$, and let $S$ be the midpoint of $CX$. Since $N, S$, and $R$ are the respective perpendicular feet from $O$ to $BC, CX$, and $PR$, we have that quadrilaterals $PRNO$ and $CNSO$ are cyclic. Furthermore, $P, S$, and $O$ are collinear as $PC = PX$. Since $ABCX$ is also cyclic, we have that
$$\angle QAB = \angle XCB = \angle PON = 180^\circ - \angle NRP = \angle MNR.$$
Analogously, we have that $\angle ABQ = \angle RMN$, so triangles $QAB$ and $RNM$ are similar. $\square$

Let $d(Z, \ell)$ denote the perpendicular distance from the point $Z$ to the line $\ell$. Using that $PR \parallel AB$ along with the similarities $QAB \sim RNM$ and $PAB \sim PMN$, we have that
$$\frac{d(Q, AB)}{AB} = \frac{d(R, MN)}{MN} = \frac{d(P, MN)}{MN} = \frac{d(P, AB)}{AB},$$
which implies that $PQ \parallel AB$.

Let $BD$ and $AC$ intersect at $T$ and let the line through $P$ parallel to $AB$ intersect $BD$ at $V$. Next, let $Q'$ be the foot of the perpendicular from $T$ to $PV$. Finally, let $Q'A$ intersect circle $ABC$ again at $X'$ and $Q'B$ intersect circle $ABD$ again at $Y'$.

Claim. $PQ'$ bisects $\angle BQ'D$ externally.

Proof. Let $PT$ intersect $CD$ at $L$. Let $\infty_{CD}$ be the point at infinity on line $CD$. From the standard Ceva-Menelaus configuration we have $(D, C ; L, \infty_{CD})$ is harmonic. Hence projecting through $P$ we have
$$-1 = (D, C ; L, \infty_{CD}) = (D, B ; T, V).$$
As $(D, B ; T, V)$ is harmonic, and also $\angle VQ'T = 90^\circ$ (by construction), the claim follows. $\square$

Now as
$$\angle Q'PD = \angle BAD = 180^\circ - \angle DY'B = 180^\circ - \angle DY'Q'$$
we have $Q'PDY'$ cyclic. By the claim, we have that $P$ is the midpoint of $\operatorname{arc} \widetilde{DQ'Y'}$, so $PD = PY'$.

Since $Y$ is the unique point not equal to $D$ on circle $ABD$ satisfying $PD = PY$, we have $Y' = Y$.
Likewise $X' = X$ so $Q' = Q$ and we are done.

Let $AX$ intersect circle $PCX$ for the second time at $Q'$. Then
$$\angle AQ'P = \angle XQ'P = \angle XCP = \angle XCB = 180^\circ - \angle BAX = \angle Q'AB$$
so $PQ'$ is parallel to $AB$. Hence, it suffices to show that $Q'$ is equal to $Q$. To do so, we aim to show the common chord of circles $PCX$ and $PDY$ is parallel to $AB$, since then by symmetry $Q'$ is also the second intersection of $BY$ and circle $PDY$.

Let the centres of circles $PCX$ and $PDY$ be $O_X$ and $O_Y$, respectively. Let the centres of circles $ABC$ and $ABD$ be $O_C$ and $O_D$, respectively.

Note $P, O_X$, and $O_C$ are collinear since they all lie on the perpendicular bisector of $CX$. Likewise $P, O_Y$, and $O_D$ are collinear on the perpendicular bisector of $DY$. By considering the projections of $O_X$ and $O_C$ onto $BC$, and $O_Y$ and $O_D$ onto $AD$, we have
$$\frac{PO_X}{PO_C} = \frac{\frac{PC}{2}}{\frac{PB + PC}{2}} = \frac{\frac{PD}{2}}{\frac{PA + PD}{2}} = \frac{PO_Y}{PO_D}.$$
Hence $O_XO_Y$ is parallel to $O_CO_D$, which is perpendicular to $AB$ as desired.