MathNet 2012 Question 23

Let $P$ be a point on the diagonal $BD$ of parallelogram $ABCD$ such that $\angle PCB=\angle ACD$. The circumcircle of triangle $ABD$ meets the line $AC$ at $E$ and $A$. Prove that
$$\angle AED=\angle PEB.\qquad\text{(Marko Đikić)}$$