4. As shown in the figure, on a semicircle with center $C$ and diameter $M N$, there are two different points $A$ and $B$. Point $P$ is on $C N$, and $\angle C A P = \angle C B P = 10^{\circ}$. If $\overparen{M A} = 40^{\circ}$, then $\overparen{B N}$ equals . $\qquad$
4. $20^{\circ}$.

From the given information, we can obtain $\angle A P C=30^{\circ}$.
In $\triangle A C P$, we have $\frac{\sin 10^{\circ}}{C P}=\frac{\sin 30^{\circ}}{A C}$.
In $\triangle B C P$, we have $\frac{\sin 10^{\circ}}{C P}=\frac{\sin \angle C P B}{B C}$.
Since $A C=B C$,
Therefore, $\sin 30^{\circ}=\sin \angle C P B=\frac{1}{2}$.
Also, $\angle C P B \neq \angle C P A$, so $\angle C P B=150^{\circ}$.
Thus, $\angle B C P=20^{\circ}$. Hence, $\overparen{B N}$ equals $20^{\circ}$.