Let $A(0,0),B(b,0),C(b,b),D(0,b)$. Choose $N(\lambda,0)$. Condition $CN=PN \Rightarrow P(0,\sqrt{2b(b-\lambda))}$. $\tan \alpha=\tan \widehat{BCN}=\frac{b-\lambda}{b}$. Slope of the line $PN\ :\ m_{PN}=-\frac{y_{P}}{\lambda}$, slope of the line $PQ\ :\ m$. Equals angles $\frac{b-\lambda}{b}=\frac{-\frac{y_{P}}{\lambda}-m}{1-m\frac{y_{P}}{\lambda}}$, calculating $m=\frac{b\lambda+by_{P}-\lambda^{2}}{by_{P}-b\lambda-\lambda y_{P}}$. Equation of $PQ\ :\ y-y_{P}=m \cdot x$. Equation of the bisector of the angle $\widehat{BQP}\ :\ mx-y+y_{P}+y \cdot \sqrt{m^{2}+1}=0$. I have checked, point $C$ lies on this bisector.

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See https://artofproblemsolving.com/community/q1h2029884p14327383

Let $T$ be the feet of an altitude from $C$ to $QP$. $\angle CPT = \angle QPN + \angle CPN = \angle BCN + \angle NCP = \angle BCP = \angle CPD$. Then $\triangle TPC \cong \triangle DPC$ and $CT = CD$. $PQ$ is tangent to a circle with the center at $C$ and radius $CB$. Therefore $\angle BQC = \angle CQP$. $2 \angle BCQ = 180 - 2 \angle BQC = 180 - \angle BQP = \angle AQP$, and we are done.