A Cute Parallelogram

Translate $\triangle APD$ by adding $\vec{AB}$ to get $\triangle BP'C \cong \triangle APD$ such that $PP' \parallel AB$ .
Let $E$ be the intersection of $PP'$ and $BC$ .
Then $m\angle CPB + m\angle BP'C = m\angle CPB + m\angle APD = 180^\circ$ , so $BP'CP$ is cyclic.
This implies
$m\angle PAD + m\angle PCB = m\angle P'BC + m\angle PP'B$ ( $BP'CP$ cyclic)
$= 180^\circ - m\angle P'EB$ (angles of a triangle sum to $180^\circ$ )
$= 180^\circ - m\angle ADC$ ( $PP' \parallel AB \parallel CD$ )
$= 180^\circ - 75^\circ = \boxed{105^\circ}$ .

Let $P'$ be the translation of point $P$ by $\vec{AB}$ . Since $APD$ gets translated to $BP'C$ , we have that $\angle BP'C + \angle CPB = 180^\circ$ , which shows that these 4 points are concyclic. Let $PP'$ intersect $BC$ at $X$ . Hence, $\angle DAP + \angle BCP = \angle CBP' + \angle BCP = \angle CPP' + \angle BCP = 180^\circ - \angle PXC$ . Since $PP'$ is parallel to $DC$ , $\angle PXC = 180^\circ - \angle XCD = \angle CDA$ , so $\angle DAP + \angle BCP = 180^\circ - 75 ^ \circ = 105 ^\circ$ .