4. $\frac{20 \sqrt{7}}{7}$.

As shown in Figure 8, draw $A F \perp B C$ at point $F$, and let $H$ be the orthocenter of $\triangle A B C$.

Obviously, points $B, C, D, E$ are concyclic, with $B C$ as the diameter.

Draw the reflection of point $E$ over $B C$ to get point $G$, which lies on the circle.
Connect $D F, P G, E F,$
$G F$. Then,
$$
E F=F G, \angle E F B=\angle G F B \text {. }
$$

Since points $C, D, H, F$ and $B, E, H, F$ are concyclic, we have
$$
\begin{array}{l}
\angle C F D=\angle C H D=\angle B H E \\
=\angle B F E=\angle G F B \\
\quad \Rightarrow D, F, G \text { are collinear. }
\end{array}
$$

Thus, $P D+P E=P D+P G$
$$
\geqslant D G=D F+F G=D F+E F \text {. }
$$

Therefore, the minimum value of $P D+P E$ is achieved when point $P$ coincides with $F$.
It is easy to find that $B F=4, E F=2 \sqrt{7}, C F=2$.
By the intersecting chords theorem, we get
$$
D F=\frac{B F \cdot C F}{F G}=\frac{B F \cdot C F}{E F}=\frac{6 \sqrt{7}}{7} \text {. }
$$

Therefore, the required minimum value is $\frac{20 \sqrt{7}}{7}$.
4. As shown in Figure $4, B D$ and $C E$ are altitudes of acute $\triangle A B C$, $\angle A B C=60^{\circ}$, $A E=8, B E=4$, and $P$ is a moving point on side $B C$. Then the minimum value of $P D+P E$ is $\qquad$.