Let $ABCD$ be just a parallelogram and $\angle APD + \angle CPB = 180^\circ$ $\Longrightarrow$ circumcircles $(O_1), (O_3)$ of $\triangle APD, \triangle CPB$ are congruent. Translating $\triangle APD$ by $\overrightarrow{AB}$ into a $\triangle CP'B$ creates a cyclic quadrilateral $PBP'C$ and parallelogram $PABP'$ $\Longrightarrow$ circumcircles $(O_2), (O_3)$ of $\triangle BPA, \triangle CPB$ are also congruent. It follows that $\angle PCD = \angle DAP,$ which means that isogonal conjugate $P^*$ of $P$ WRT $\triangle ACD$ is on perpendicular bisector of $AC.$ As a result, the locus of $P$ is a rectangular circum-hyperbola of the parallelogram $ABCD,$ centered at its diagonal intersection $E.$ If $ABCD$ is a rhombus, this hyperbola degenerates to its diagonals $AC, BD.$