<Qua-2023-719>, hint

Let O be the center of square ABCD and P on OA,

Then triangles  POB and POD are congruent and so <APB=<DPC.

Suppose P' on the extension of  BO interior of triangle OAD, then triangles P'AP and

PCP' are not congruent. <AP'B=<APB-<P'AO and let Q be the intersection of CP' and DP

then<QCP and <QDP' are not equal. and <P'AO and <P'CO are not equal.
Hence <AP'B+<DP'C are not 180.

Diagonals of AC BD are the desired locus.