a)$\angle DPC=90-\frac{\angle PDC}{2}=90-\frac{180-\angle PAB}{2}=\angle PAB=90-\angle APB$ Thus $\angle BPC=90$ b)Let $R'$ be the other intersection of the circumcircle of $\triangle BPC$ with $AD$. we have $QR=QP$. Then $\angle PRQ=\angle RPQ=\angle RPB+\angle BPQ=\angle ABP+\angle PBQ=\angle ABQ$ from which $ABQR'$ is concyclic. Thus $R'=R$ and the conclusion follows.

Alternative solution: a) Same. b) Notice that PQ is a median to a hypotenuse. Hence $PQ=\frac{BC}{2}=BQ \longrightarrow BPQ$ is isosceles. Since $APB$ is also isosceles, our quadrilateral $APBQ$ is a kite. Hence $AQ$ is the angle bisector of $\angle PAB$. Hence $\angle QRB = \angle QBR \longrightarrow RQ=QB=QC$. Hence RQ is a median of triangle CRB that is half the length of the hypotenuse $BC$. Thus $\angle CRB = 90= CPB$, and thus $P, C, R, B$ are cyclic by congruent inscribed angles.

$PB, PC$ are perpendicular onto the bisectors of $\angle BAD, \angle ADC$ respectively, hence perpendicular onto each other. $ABQP$ is a kite, hence $\angle ABQ+\angle DPQ=180^\circ \ (\ 1\ ), ABQR$ is cyclic, $\angle ABQ+\angle ARQ=180^\circ \ (\ 2\ )$, hence $\angle PRQ=\angle RPQ, \Delta PRQ$ is isosceles, $PQ=RQ$, hence $B, C, P, R$ are equally apart from $Q$. Best regards, sunken rock

For part (b): easy to see, that, $QD \perp PC$ and $AQ \perp BP$ $\implies$ $\angle QRB=\angle QAB=\frac{1}{2} \angle A$ $=$ $\frac{1}{2} (180^{\circ}-D)$ $=$ $90^{\circ}-\angle CRQ$ $\implies$ $\angle BPC$ $=\angle BRC=90^{\circ}$ implies the desired result