Let $F$ be a point out of the parallelogram $ABCD$ such that $ \triangle AFD = \triangle BPC$ because $AD||BC$ $PCDF$ and $PBFA$ are parallelograms and so $PB=AF$ and $PC=FD$ So $AD\cdot BC = PD \cdot BE + PA \cdot FD$ In other words $PA \cdot PC + PB \cdot PD = AB \cdot BC$ (Opposite of Ptolemy) Thus, $AEFD$ is cyclic $(1)$ Angle chasing gives $\angle PBC =\angle FAD=\angle FPD= \angle PDC $ $(2)$ Let now $X,Y,Z,W$ be the projections of $P$ to $AB,BC,CD,DA$ Its trivial that $X,P,Z$ and $Y,P,W$ are collinear and now $XPBY$ $PYZC$ $PZWD$ $PWAX$ are concyclic $\angle PBC= \angle PXY$ $\angle PDC= \angle PWZ$ Therefore, $\angle PXY=\angle PYZ$ So $XYZW$ is concyclic Notice from power of point that $PZ \cdot PX=PW \cdot PY$ multiply by 4 we get the same result for $(2PX,2PY,2PZ,2PW) $ (The reflection is twice the projection) so the reflections of $P$ over lines $AB$, $BC$, $CD$, and $DA$ are concyclic.