4. In isosceles $\triangle A B C$, $A B=A C=6, P$ is a point on side $B C$, and $P A=4$. Then $P B \cdot P C$ equals: ( ).
(A) 10
(B) 15
(C) 20
(D) 25
4. (C).

As shown in the figure, on the extensions of $AC$ and $BA$, intercept $AD = AE = AP = 4$, and connect $PD$ and $PE$. Then $\angle APD=$
$$
\begin{array}{l}
\angle ADP, \angle APE=\angle E, \angle BAP \\
=2 \angle E . \\
\begin{array}{l} 
\because \angle APD= \angle APC-\angle CPD \\
=\angle B+\angle BAP-\angle CPD \\
=\angle B+2 \angle E-\angle CPD, \\
\angle ADP= \angle C+\angle CPD, \\
\therefore \angle B+2 \angle E-\angle CPD=\angle C+\angle CPD .
\end{array}
\end{array}
$$

Given $\angle B=\angle C$, we know $\angle E=\angle CPD$.
Thus, $\triangle BEP \backsim \triangle CPD$.
Therefore, $PB \cdot PC = BE \cdot CD$
$$
=(AB \div AP)(AC - AF) = 20 .
$$