<Qua-2025-762>, hint

Let Q be midpoint of AN, then  since ABN is isosceles <AQB=90 .

<BAQ=<BNQ=<BCP=<QBC, Hence <APL=90. Since  triangles ACM and ABN are 

congruent and BQ //PL,  L is midpoint of DM

Triangles LDP and LMP are congruent, DP=PM.

<CML+<CMB=<BAQ+<MBN. 2<MBN+2CML+90=180.

Hence <MBN+<CML=45,  <BMC+<CML=45 (<MBN=<BMC). Hence <DPM=90.