All angles in the following proof are directed; I'm not sure the latex symbol for directed angles so the symbol for regular angles will have to do. Let $O$ be the midpoint of $PQ$; it is immediate that $O$ is the circumcenter of $APBQ$. Notice that $\angle OBP = \angle BPQ = \angle BAQ = 90^\circ - \angle PAB = \angle DAP = \angle PAX$; similarly, $\angle OAP = \angle PBX$. Therefore, $\angle OAX = \angle OBX$, obtaining $ABOX$ concyclic. From here, $\angle QOX = \angle QOB + \angle BOX = \angle BAX + 2 \angle OQB = 90^\circ + 2 \angle DAP + 2 \angle PAB = 90^\circ \pm 180^\circ = 90^\circ$, yielding $OX$ is the perpendicular bisector of $PQ$ and the result follows.

One - liner Let $\omega$ be the circle with diameter of $PQ$ and $A_1, B_1$ be the intersection of $\omega$ and $BC$, $AD$, and $A_2, B_2$ be the symmetrical point of $A_1,B_1$ with respect to $PQ$. $X$ is the intersection of $AB_2$ and $BA_2$, and the center of $\omega$ ; $O$, is the intersection of $AA_1$ and $BB_1$. We can see that $A_1A_2$ and $B_1B_2$ is perpendicular to $PQ$. Apply Pascal on six points $\left ( \begin{matrix} A & A_2 & B_1 \\ B & B_2 & A_1 \end{matrix} \right )$, and we get that $X,O$ is perpendicular to $PQ$, which means that $PX = QX$.

Not the easiest or cleanest solution but the the first one I found: Construct $E$ and $F$ along $AD$ such that $AF=AX=AE$ and $E$ is closer to $D$ than $F$. Construct $H$ and $G$ along $AD$ such that $BG=BX=BH$ and $G$ is closer to $C$ than $H$. It is then immediate that $PE=PX=PG$ and $QF=QX=QH$. Notice that we must have $\angle APB<90^{\circ}$. Thus $$\angle EPG=2\angle APB=180^{\circ}-2\angle AQB=\angle FQH$$Combined with the fact that $FH=EG$ we get that $FQH$ and $EPG$ are congruent isosceles triangles. Thus $FQ=EP\Rightarrow PX=QX$, as desired.

Attachments:

Because $\angle ABC = \angle QBP$, $\angle QBA=\angle PBC=\angle PBX$. Similarly $\angle XAP=\angle QAB$. So $X$ has an isogonal conjugate in quadraliteral $AQBP$, thus $\angle QXB + \angle AXP=180$. By the Law of Sines in triangles $QXB$ and $AXP$, $\frac{QB}{AP}=\frac{QXsinQAB}{XPsinABP}$, and as $\frac{QB}{AP}=\frac{sinQAB}{sinABP}$, $QX=XP$

This is just an angle chase, given the right reference. Will post solution later, PM me if I don't.