Let the line passing through $B$ and parallel to $PC$ intersect the line $AE$ at $F$ and $CQ$ at $F'$. Let $QD$ and $BF$ intersect at $K$. Since the triangles $PEC$ and $FEB$ are similar, $PD = DC$ and $D, E, K$ are collinear, we obtain that $K$ is the midpoint of the line segment $BF$. On the other hand, since the triangles $QPC$ and $QBF'$ are similar, $PD = DC$ and $Q, D, K$ are collinear, we also obtain that $K$ is the midpoint of the line segment $BF'$. Therefore the points $F$ and $F'$ must coincide. Since $\angle EAC = \angle PCE = \angle CBF$, we conclude that the points $A, B, F, C$ are concyclic and hence $\angle BAF = \angle BCF = 180^\circ - \angle BCQ$ which finishes the proof.