3. As shown in Figure 2, in $\triangle A B C$, $A B=A C$, $\angle B A C=78^{\circ}$, $P$ is a point inside $\triangle A B C$, and $\angle B C P=42^{\circ}$, $\angle A P C$ $=162^{\circ}$. Then the degree measure of $\angle P B C$ is
3. $21^{\circ}$.

As shown in Figure 7, construct an equilateral $\triangle APO$ on side $AP$ at point $B$, and connect $BO$.
It is easy to see that
$$
\begin{array}{l}
\angle ABC = \angle ACB \\
= 51^{\circ}.
\end{array}
$$

Then $\angle PCA = \angle ACB - \angle BCP$ $= 51^{\circ} - 42^{\circ} = 9^{\circ}$.
Thus, $\angle PAC = 180^{\circ} - \angle APC - \angle PCA$ $= 180^{\circ} - 162^{\circ} - 9^{\circ} = 9^{\circ}$.
Also, $\angle OAB = \angle BAC - \angle PAC - \angle PAO$ $= 78^{\circ} - 9^{\circ} - 60^{\circ} = 9^{\circ}$,
so $\angle OAB = \angle PAC$.
Since $AB = AC$ and $AO = AP$, we have $\triangle OAB \cong \triangle PAC$.
Thus, $\angle ABO = \angle ACP = 9^{\circ} = \angle OAB$.
Therefore, $OB = OA = OP$.
Hence, $O$ is the circumcenter of $\triangle ABP$.
So, $\angle ABP = \frac{1}{2} \angle AOP = 30^{\circ}$.
Therefore, $\angle PBC = \angle ABC - \angle ABP = 21^{\circ}$.