Let $\angle{CBA}=2\alpha$. Since $EF\parallel AB$ and $CE=DE=BE$, we have $$\boxed{\angle{CEP}=\angle{PED}}=2\alpha$$ Also $\angle{PAF}=45^o-\alpha$ and $\angle{PFE}=90^o-\angle{FEC}=90^o-2\alpha$, so $\angle{FPA}=45^o-\alpha$. $\Rightarrow \angle{FAP}=\angle{APF}$ which means $$FA=FP=FC\Rightarrow \angle{APC}=90^o\Rightarrow \angle{FPC}=90^o-(45^o-\alpha)=45^o+\alpha\Rightarrow \angle{CEP}=2\alpha\Rightarrow \boxed{\angle{PCE}=45-\alpha}$$ From $\angle{ACD}=2\alpha$ we finally conclude $$\boxed{\angle{DCP}=\angle{PCE}}=45^o-\alpha$$ So $P$ is the incenter of the triangle $CDE$.

Lemma 1: $ADPC$ is cyclic Let's call $\angle BCP=\beta$ and $\angle PCD=\alpha$. Note that: $FD=FA=FC$ , those three are equal to to $FP$, because $\angle FAP=\angle FPA=\dfrac{\alpha + \beta}{2}$, because $\angle CFP=\alpha+\beta$, which comes from $EDFC$ being cyclic, because $EF$ is the perpendicular bissector of $CD$ and $CE=DE$. So the Lemma 1 was proved because there is a circle that passes through $CPDA$ with center $F$. But $BC^2=BD.BA$, by similar triangles, which tells us that $BC$ is tangent to the circle circunscribed in $CPDA$, which finishes the proof because then $\dfrac{\alpha+\beta}{2}=\beta$, thus $\alpha=\beta$. And then $P$ is the incenter of triangle $DCE$.