Angle Outlier

$m\angle BOC=2\cdot m\angle BAC=160°$ , thus $m\angle BCO=\frac{1}{2}(180°-m\angle BOC)=10°$ . $m\angle PCO=m\angle CPO$ and due to parallelism, $m\angle CPO=m\angle BCP$ , then $m\angle PCO=m\angle BCP=\frac{1}{2}m\angle BCO=5°$ , leading to the answer that $m\angle CPO=5°$ .

<(boc)=160 and <obc=10 then< pob=10 thats is arc (pB) so <pcb =5 and< p =5

Joining $PO$ to meet the circle at $X$ , and joining $BX, PB, PA$ , we get $\angle PAB = \angle PXB = \angle PCB = \angle XPC = \theta$ , where the first two equalities follow from them being angles in the same segment and the last one follows from parallelity of $OP$ and $BC$ .

Also $PX$ being the diameter, we get $\angle PBX = 90^{\circ} = \angle PXB + \angle BPX = \theta + \theta + 80^{\circ} \implies \theta = 5^{\circ}$