smc

Since $\angle CAP=\angle CBP=10^{\circ }$, quadrilateral $ABPC$ is cyclic (as shown below) by the converse of the theorem "angles inscribed in the same arc are equal". Since $\angle ACM=40^{\circ }$, $\angle ACP=140^{\circ }$, so, using the fact that opposite angles in a cyclic quadrilateral sum to $180^{\circ }$, we have $\angle ABP=40^{\circ }$. Hence $\angle ABC=\angle ABP-\angle CBP=40^{\circ }-10^{\circ }=30^{\circ }$. Since $CA=CB$, triangle $ABC$ is isosceles, with $\angle BAC=\angle ABC=30^{\circ }$. Now, $\angle BAP=\angle BAC-\angle CAP=30^{\circ }-10^{\circ }=20^{\circ }$. Finally, again using the fact that angles inscribed in the same arc are equal, we have $\angle BCP=\angle BAP=\boxed{20^{\circ }}$.