My solution: Easy to prove $\triangle AFC =\triangle CEB \Longrightarrow \angle ACF =\angle EBC \Longrightarrow AC$ is the tangent of circumcircle described $\triangle BPC$ By angle changing we can get $AB$ is the tangent of circumcircle described $\triangle BPC$, too. Let $AP \cap (BPC) =L \Longrightarrow BPCL$ is a harmonic quadrilateral. By Plotemy theorem, we get $LP.BC=LC.BP+LB.CP=2LC.BP.$ $\Longrightarrow \frac{LP}{2LC} =\frac {BP}{BC} . \Longrightarrow \frac{LP}{LC} =\frac {BP}{BD} \Longrightarrow \triangle BDP \sim \triangle LCP$ $\Longrightarrow \angle BPD =\angle LPC =\angle APF$ Q.E.D

Nice solution with Ptolemy Since $AB$ and $AC$ are tangents to the circumcircle of triangle $BPC$, one can prove this easily with angle chasing and stuff. Now, it is a well-known fact that $AF$ coincides with a symmedian of $ABC$. And we are done

Let $F_1$, $P_1$ be the images of $F$, $P$ under the rotation about $C$ that takes $A$ to $B$ and let $PD$ cut $CF_1$ at $Q$. We easily get that $BECF_1$ is a parallelogram, and that $\triangle BF_1Q \cong \triangle CEP$, so $BPP_1Q$ is an isosceles trapezium. Hence $\angle BPD = \angle BPQ = \angle BP_1P = \angle BP_1F_1 = \angle APF$. Done.

@Juckter, do you remember the official solution for this problem?

$AB, AC$ are tangent to the circle $BCP$, so $AP$ is symmedian, done (seen before on the site). Best regards, sunken rock