<Qua-2025-799>, hint

AC and BD intersect at Q.Let QA=a, QB=b, QC=c, QD=d

Since KP//BC,  QK/QP=QB/QC, hens QK=tb, QP=tc 

Since triangles QAK and QCM are similar QK=xa QM=xc

QK=tb=xa---> x/t=b/a  QP/QM=t/x=a/b. Thus triangles QPM and QAB are similar

<QPM=<QAB. <QPM+QPK=<QAB+<QCB=90