4. Let the side length of the equilateral $\triangle ABC$ be $2, M$ is the midpoint of side $AB$, $P$ is any point on side $BC$, and $PA + PM$ are denoted as $s$ and $t$ for their maximum and minimum values, respectively. Then $s^{2}-t^{2}=$ $\qquad$.
$4.4 \sqrt{3}$.
First, find $s$.
$$
\begin{array}{l}
\because P A \leqslant A C, P M \leqslant C M, \\
\therefore P A+P M \leqslant C A+C M=2+\sqrt{3} .
\end{array}
$$

When point $P$ is at vertex $C$, the equality holds.
Thus, $s=2+\sqrt{3}$.
Next, find $t$. As shown in Figure 4, construct the equilateral $\triangle A^{\prime} E C$. Let $M^{\prime}$ be the midpoint of $A^{\prime} B$, then $\triangle P B M \cong \triangle P B M^{\prime}$.
$$
\begin{array}{c}
\therefore P M=P M^{\prime} . \\
P A+P M=P A+P M^{\prime} \geqslant A M^{\prime} .
\end{array}
$$

Connect $C M^{\prime}$, then $\angle A C M^{\prime}=90^{\circ}$.
$$
\begin{array}{l}
\therefore A M^{\prime}=\sqrt{A C^{2}+\left(C M^{\prime}\right)^{2}}=\sqrt{4+3}=\sqrt{7} . \\
\therefore t=\sqrt{7} .
\end{array}
$$

Therefore, $s^{2}-t^{2}=(2+\sqrt{3})^{2}-7=4 \sqrt{3}$.