Let $Q'$ be a point in $AD$ such that $CD^2=BD^2=DQ'.DA$ $\Longrightarrow$ $\measuredangle Q'CD=\measuredangle DAC=\alpha$ and $\measuredangle DBQ$ $=$ $\measuredangle DAB$ $=$ $\alpha+\beta$. Let $P'$ be a point in $QC$ such that $\measuredangle BP'C=90^{\circ}$ $\Longrightarrow$ $DB=DC=DP'$ $\Longrightarrow$ $\measuredangle DAC=\measuredangle DP'C=\alpha$ $\Longrightarrow$ $DP'AC$ is cyclic hence $\measuredangle DAC$ $=$ $\measuredangle P'AD$ $=$ $\alpha...(1)$ and since $\measuredangle BP'C=90^{\circ}$ we get $\measuredangle P'BD=90^{\circ}-\alpha...(2)$. On the other hand it so easy note that $P$ is unique $\Longrightarrow$ $Q$ is unique $\Longrightarrow$ by $(1)$ and $(2)$ we get $P'=P$ and $Q'=Q$ hence $\measuredangle BQP=\measuredangle BAC=2\alpha+\beta$.

It is clear that $P$ is unique. Let the line which is reflection of $AC$ across $AD$ cut $BC$ at $E$. Let $(ADC)\cap AE = P'$. Note that $\angle P'CD = \angle DP'C \implies DC=DP'=DB \implies \angle BP'C = 90 \implies \angle P'BD = 90 - \angle CAD \implies P'= P$. So we have $\triangle AQC \sim \triangle ACD \implies DC^2 = DQ\cdot DA = DB^2 \implies \triangle AQB \sim \triangle ABD$. So we have $\angle QBD = \angle BAD$ and $\angle QCB = \angle CAD \implies \angle PQB = \angle BAC$. $\square$

Let $A'$ be the reflection of $A$ to $D$ and $CQ \cap AB=R,BW\cap AC=S,AP\cap A'B=M$ $\angle A'AM=\angle DAP=\angle CAD= \angle MA'A \implies MA=MA'$ $MA=MA'$ and $AD=A'D \implies MD \perp AA'$ $\angle PMD=\angle AMD=90-\angle DAM=\angle PBD \implies P,B,M,D$ are cyclic. $\angle BDP=\angle BMP=2\angle DAP=\angle CAP \implies A,P,D,C$ are cyclic. $\angle CPD=\angle CAD=\angle DAP=\angle DCP \implies DB=DC=DP \implies BP \perp PC$ $\angle C= \angle QCA+\angle DCP=\angle QCA+\angle CAQ=\angle CQD=\angle RQA=\angle RSA \implies R,Q,S,A$ are cyclic. $\angle BQP=\angle BQR=\angle CAB$ as desired.

Let $X$ be the reflection of $P$ wrt $AD$.From angle condition,we get $BPXC$ is cyclic.So $AD$ goes through midpoints of both $PX$ and $BC$,and is the perpendicular bisector of $PX \implies D$ is the circumcenter.$\angle QAX=\angle PXB=\angle PCB=\angle QCB$,so $Q$ is the $A$ humpty point of $\triangle {ABC}$.$\angle PQB=\angle QBC+\angle QCB=\angle BAD+\angle DAC=\angle BAC$.