Consider the point $X$ on the segment $BC$ such that $AX=BC$. Since $\angle CBA= \angle CDA = 90^{\circ}$, $ABCD$ is cyclic. Thus, $\angle CAD=\angle CBD$. Furthermore, $AX=BC\land AD=BP$, so triangles $AXD$ and $BCP$ are congruent. Let $\angle DBA=2\alpha$. Since $XCD$ is isosceles, $\angle XCD=90^{\circ} - \alpha=180^{\circ} - \angle AXD=180^{\circ} - \angle BCP$, but $\angle CBK=90^{\circ} - \alpha$, so we are done.