10. Given point $P$ inside $\triangle A B C$, satisfying $\angle A B P$ $=20^{\circ}, \angle P B C=10^{\circ}, \angle A C P=20^{\circ}$ and $\angle P C B$ $=30^{\circ}$. Then $\angle C A P=$
10.20.

As shown in Figure 7, construct a regular $\triangle QBC$ on side $BC$ on the same side as point $A$.

Notice that $BA$ is the angle bisector of the regular $\triangle QBC$, so, $\angle AQC = \angle ACQ = 10^{\circ}$.

Also, $CP$ is the angle bisector of the regular $\triangle QBC$, hence
$$
\angle PQC = \angle PBC = 10^{\circ}.
$$

Since points $A$ and $P$ are on the same side of $QC$, point $A$ lies on segment $PQ$.
Thus, $\angle CAP = \angle AOC + \angle ACQ = 20^{\circ}$.