Solution:

The case when $AD \parallel BC$ is simple. Namely, in that case $E$ is the midpoint of the diagonal $AC$, so $EA = EC = EP$, and similarly $EB = ED = EQ$ holds as well. From this it follows that the circles $CEQ$ and $DEP$ are symmetric with respect to the perpendicular bisector of the segments $CP$ and $DQ$, so $EF \perp CP$, i.e. $EF \parallel AP \parallel BQ$.

Let us denote by $O$ the point of intersection of the lines $AD$ and $BC$. Triangles $OAP$ and $OBQ$ are similar, so $\frac{OP}{OQ} = \frac{OA}{OB} = \frac{OD}{OC}$, i.e. $OC \cdot OP = OD \cdot OQ$, from which it follows that the points $C, D, P$ and $Q$ lie on some circle $\gamma$.

Let the lines $AP$ and $BQ$ intersect at point $H$, and the lines $DP$ and $CQ$ at point $G$. Since $GC \cdot GQ = GD \cdot GP$, the point $G$ has equal power with respect to the circles $CEQ$ and $DEP$, so it lies on their radical axis $EF$. On the other hand, the points $E, G$ and $H$ are collinear on the

basis of Pappus's theorem for the triples of points $B, C, P$ and $A, D, Q$. It follows that all four points $H, E, G$ and $F$ are collinear.