13. As shown in Figure 4, in rhombus $A B C D$, $A E \perp B C$ at point $E$. If $\cos B$
$$
=\frac{4}{5}, E C=2 \text {, }
$$
$P$ is a moving point on side $A B$,
then the minimum length of line segment $P E$ is $\qquad$
13.4.8.

Let the side length of square $ABCD$ be $x$. Then $AB=BC=x$.
Given $EC=2$, we know $BE=x-2$.
Since $AE \perp BC$ at point $E$, in the right triangle $\triangle ABE$, $\cos B=\frac{x-2}{x}$.
Given $\cos B=\frac{4}{5}$, we have $\frac{x-2}{x}=\frac{4}{5}$.
Solving for $x$ gives $x=10$, so $AB=10$.
It is easy to find $BE=8, AE=6$.
When $EP \perp AB$, $PE$ reaches its minimum value.
Thus, by the triangle area formula, we have
$$
\frac{1}{2} AB \cdot PE=\frac{1}{2} BE \cdot AE \text {. }
$$

Solving for the minimum value of $PE$ gives 4.8.