I think it's supposed to be $\angle{B} = \angle{D} = 90^{\circ}$. $K$ is the reflection of $B$ over $AD$. If $X = AP \cap CE$, then it suffices to show that $PBXE$ is cyclic, since this would imply $\angle{EBX} = \angle{EPX} = \angle{KFA} = \angle{AFB} = \angle{BEX}$. However, we know that $\angle{PBE} = \angle{PAE} - \angle{AFB} = \angle{PXE}$, so $PBXE$ is cyclic. $\square$

Solved with mathscrazy and Rawlat_vanak First, observe that $F$ is orthocentre of $AEC$. We will prove that $P$ is $A$-humpty point of $AEC$. Clearly this is sufficient. To prove this, first notice that it lies on circle with diameter $AF$. Also, angles give $$\angle CEA = \angle ABK = \angle APE$$These two conditions force that $P$ must be the Humpty point, and in particular AP bisects CE. Done! @Starchan, beat you

$CA\cap EF=G\in (ABF)$ since $A$ is the orthocenter of $FEC$. $\angle FPK=180-\angle KBF=180-\angle ECF$ thus, $P$ is the $F-$queue point of $FEC$ which gives the desired result.$\blacksquare$