Example 3 As shown in Figure 1, in $\triangle A B C$, $A B=A C=$ $4, P$ is a point on side $B C$ different from points $B, C$. Then the value of $A P^{2}+B P \cdot P C$ is ( ).
(A) 16
(B) 20
(C) 25
(D) 30
Solution 1 As shown in Figure 1, draw $A D \perp B C$.
Since $A B=A C$, we have $B D=D C$.
$$
\begin{array}{l}
\text { Therefore, } A P^{2}+B P \cdot P C \\
=A D^{2}+P D^{2}+(B D-P D)(D C+P D) \\
=A D^{2}+P D^{2}+B D^{2}-P D^{2} \\
=A B^{2}=16 .
\end{array}
$$

Solution 2 Let point $P$ be the midpoint $D$ of $B C$, at this time,
$$
\begin{array}{l}
B P=P C=B D, A P=A D . \\
\text { Then } A P^{2}+B P \cdot P C=A D^{2}+B D^{2} \\
=A B^{2}=16 .
\end{array}
$$

Therefore, the answer is A.