Let the circle with diameter $AD$ (which we will denote by $(AD)$ ) intersect $(ABC)$ at $F'$. Its easy to see that the $(BDC)$ and $(AD)$ are tangent at $D$, because their centers are collinear on the line $AP$. By radical axis on $(ABC), (AD), (BDC)$, we see that $BC, AF'$ and the line through $D$ perpendicular to $AP$ intersect at one point, say $E'$. Then, we have $\angle ADE' = \angle AKE' = 90$ so $ADKE'$ is concyclic. This solves the problem.

Nice and straightforward. Since $\angle ADE = \angle AKE = 90^{\circ}$, it suffices to show $ED^2 = EF \cdot EA$. But $EF \cdot EA = EB \cdot EC$ from the circumcircle of $ABC$, so it suffices to justify $ED^2 = EB \cdot EC$, equivalent to $\angle EDB = \angle BCD$. But if $\angle BCD = x$, then $\angle BPD = 2x$ and $\angle BDP = 90^{\circ} - x$ as $DP = BP = PC$, thus $\angle EDB = 180^{\circ} - \angle ADE - \angle BDP = x$, as desired.

nice problem

$BP=PC=DP$. Then Pythagoras in $\triangle EDP$ gives $$ED^2+BP^2=EP^2=(EB+BP)^2$$Thus $$ED^2=EB(EB+2BP)=EB\cdot EC=EF\cdot EA$$as desired.