Ireland

Extend $CP$ to meet the circumference at $E$. Then, using the assumption, we get $\angle BPE=\angle APC=\angle DPB$, hence $D$ is the reflection of $E$ in the line $AB$. Therefore, $ED$ is perpendicular to $AB$ and so $\angle BPE+\angle PED=90^{\circ }$.



We also have $\angle COD=2\angle CED=2\angle PED$, as these angles are subtended by the same arc $CD$. Because $|OC|=|OD|$ the triangle $CDO$ is isosceles and we obtain $2\angle BPE=180^{\circ }-2\angle PED=180^{\circ }-\angle COD=2\angle DCO$. This implies $\angle DPB=\angle BPE=\angle DCO$, hence $PODC$ is cyclic.