$\angle CBP = 90^\circ - \angle DBP$, so let the perpendicular to $PB$ at $P$ meet $BD$ at $X$. Then if $\angle BAP = \angle CBP$, we have $\angle PXB = \angle PAB$, so $XAPB$ is cyclic. So, $\angle BAX = 90^\circ$, so $X \equiv D$. If $DAPB$ is cyclic, $\angle PAB = \angle PDB = 90^\circ - \angle DBP = \angle CBP$, so done. Note : directed angles make this solution complete, but I'm lazy enough to direct them.

https://artofproblemsolving.com/community/u417508h1666711p10592130

Consider the case when B is obtuse. Most of the contestants got only 6 points because of it