Consider the circumcircle $(L)$ of $ABCO$. Let $I,J$ the centers of $(BPX),(CYP)$; denotes $\angle BOC=2\alpha $ then $ \angle BAO=\frac{\pi}{2}-\alpha$ we claim that $(L)$ is tangent to $(BPX),(CYP)$ internally indeed: $\angle IBA=\angle IBX=\frac{\pi}{2}-\angle BPX=\frac{\pi}{2}-\alpha=\angle BAO$ thus $B,I$ and $L$ are collinear idem for $C,J$ and $L$ hence $PZ$ go through the radical center of the three circles which projectivelly exists but is infinite point if the tangents at $B,C$ are parallel which case is when $B,C,L$ are collinear i.e. $ABCO$ is a square which leads $OA=\sqrt{2} r$

Answer: $k=\sqrt{2}$ Let $D$ be the point along line $AO$ such that $\angle ABD=\angle BPC$. Notice that $D$ is the point at infinity when $\angle ABD+\angle BAD=180^{\circ}\Rightarrow$ $\frac{1}{2}\angle BAC+\angle BPC=180^{\circ}\Rightarrow \angle BAC=90^{\circ}$. It is sufficient to show that $D$ has equal powers with respect to $(BPX)$ and $(CPY)$ or that the tangents from $D$ to these circles have equal length. We claim in fact that $DB$ and $DC$ are tangent to $(BPX)$ and $(CPY)$. This is just an angle chase $$\angle PBD=\angle BPC-\angle XBP=\angle PXB$$$$\angle PCD=\angle CPB-\angle YCP=\angle PYC$$