Let $M$ be the midpoint of $BC$. Notice $\angle ADE = \angle QDB = \frac{1}{2} \angle B$. We also have $\angle DMP = \angle QBD = 90^\circ$. Then we know that $\angle BPQ = \angle BQP = 90^\circ - \frac{\angle B}{2}$, so $B$ is the intersection of the perpendicular bisector of $PQ$, and the circle with diameter $DQ$. Then, we see that $M$ is the second intersection of $BP$ and the circle with diameter $DP$. Finally, we see that $C$ is the reflection of $B$ over $M$, and $A$ is the intersection point of $DM$ and $BQ$.