Distinct points $A$ and $B$ are on a semicircle with diameter $MN$ and center $C$. 
The point $P$ is on $CN$ and $\angle CAP = \angle CBP = 10^{\circ}$. If $\stackrel{\frown}{MA} = 40^{\circ}$, then $\stackrel{\frown}{BN}$ equals

$\textbf{(A)}\ 10^{\circ}\qquad \textbf{(B)}\ 15^{\circ}\qquad \textbf{(C)}\ 20^{\circ}\qquad \textbf{(D)}\ 25^{\circ}\qquad \textbf{(E)}\ 30^{\circ}$
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 $\angle BCP = \angle BAP = \boxed{\textbf{(C)}\ 20^{\circ}}$.