Let the centres of the circumcircles of triangles $APD$ and $BQD$ are $E$ and $F$ and $G$ is point of intersection of $PQ$ and circumcircle $BQD$ $\angle QGD = \angle QBD = \angle CAD \to \angle PGD = \angle PAD \to $ point $G$ lies on circumcircle of $\triangle APD$ $\angle ADG = \angle APG=90^o \to DG \perp AD$ $DG$ is common chord of the circumcircles of triangles $APD$ and $BQD$ so $EF \perp DG \perp AD \to EF \parallel AD$

diagram please?

I solved this by spotting $2$ more cyclic quadrilaterals. I am not posting full solution but rather main ideas of construction: Let $O_{1}$ and $O_{2}$ be the circumcenters of triangle $APD$ and triangle $BQD$ respectively. Now let $O_{1}P \cap O_{2}Q=L$ then, $DO_{1}O_{2}L$ and $PQLD$ are cyclic quadrilaterals. Then finish by angle chasing.

Attachments:

We consider the $A-$antipode w.r.t $(APD)$ as $E$, also since $$\angle DBQ=\angle DAC=\angle DEP$$$(BQDE)$ is cyclic, so the perpendicular bisector of $ED$ will pass through $O_1,O_2$ and since the perpendiular bisector will be parallel to $AD$ we are done.