We only need to prove that $PE$ passes through the circumcentre of $\Delta PAB.$ Since $\angle DPE + \angle PAB=90$ And we are done

It's clear that, $\Delta APD \sim \Delta BPC$ and $\Delta APB \cong \Delta DPC \implies ABCD \text{ is an isosceles trapezium} \implies \text{Cyclic Quadrilateral}$, Let $\angle BAP=A$ and $\angle APB=B$ and Let, $AP=b ; BP=a$, Also, Let $PE \cap AB=K$ Therefore, after some simple angle chasing, we arrive at, $\angle BPK =90^{\circ}-A \text{ and } \angle APK=90^{\circ}-B$ Now, $$\frac{AK}{KB}=\frac{\tfrac{1}{2}b \cdot PK \sin (90^{\circ}-B)}{\tfrac{1}{2}a \cdot PK \sin (90^{\circ}-A)}=\frac{b\cos B}{a\cos A}$$By Sine Rule, $$\frac{b}{\sin B}=\frac{a}{\sin A} \implies \boxed{\frac{b}{a}=\frac{\sin B}{\sin A}}$$Therefore, $\boxed{\frac{AK}{KB}=\frac{\sin 2B}{\sin 2A}}$ If $A\equiv (1:0:0) ; B\equiv (0:1:0) ; C\equiv (0:0:1) \implies \boxed{K\equiv (\sin 2A: \sin 2B:0)}$, This implies that cevian $PK$ passes through the circumcenter with coordinates $(\sin 2A:\sin 2B:\sin 2C)$ Therefore, Line $PE$ passes through the circumcenter which also the intersection of the perpendicular bisectors of $PA$ and $PB$ , Hence, all the three are concurrent at the circumcenter.

denote the circumcentre of PAB as O. Inversion at P（denote X become X for we do not need old points） make O be the reflection of P through AB and E the intersection of the the perpendicular from D to PD and the perpendicular from C to PC. We need to show PEO are on the same line. m(EPC)=m(EDC)=90-m(PDC)=90-m(PAB)=m(OPC) Hence we are done.

We have that $ABCD$ is isosceles trapezium, its diagonals intersect at $P$. Let $F$ be the circumcentre of $ABP$. Let $G$ be the intersection of perpendicular bisectors of $BP$ and $BC$. Thus, $G$ is the centre of $(BPC)$. We have that $\angle BFG=\frac{\angle BFP}{2}=\angle BAP=\angle BAC=\angle BDC$. Also, $\angle DBC=\angle ACB=\angle PCB=\frac{\angle PGB}{2}=\angle FGB$, hence $\triangle FBG\sim \triangle DBC$, thus $\angle FBG=\angle DCB$. We conclude that $$\angle FPG=\angle FBG=\angle DCB=180^{\circ}-\angle EPG\implies \text{$E,P$ and $F$ are collinear.}$$