Team Selection Test for JBMO

Let $Q$ be the foot of the perpendicular from $B$ to the line $AC$. Then we obtain $BQ=EQ=ED$ which means that $\angle AQD=22.5^{\circ }$. Since the points $B,F,Q,A$ are concyclic, we have $CF\cdot CB=CQ\cdot CA$. Since the points $B,F,P,D$ are also concyclic we have $CF\cdot CB=CP\cdot CD$ and hence $CQ\cdot CA=CP\cdot CD$ and therefore the points $A,Q,P,D$ are concyclic. So, $\angle APD=\angle AQD=22.5^{\circ }$.