4. In quadrilateral $A B C D$, $\angle A=60^{\circ}, A B=1$, point $E$ is on side $A B$ such that $A E: E B=2: 1, P$ is a moving point on diagonal $A C$. Then the minimum value of $P E+P B$ is $\qquad$ .
4. $\frac{\sqrt{7}}{3}$.

Since quadrilateral $A B C D$ is a rhombus, point $B$ and $D$ are symmetric with respect to diagonal $A C$. Therefore,
$$
P E+P B=P E+P D \geqslant D E \text {. }
$$

Equality holds if and only if $P$ is the intersection of $D E$ and $A C$.

Thus, the minimum value of $P E+P B$ is the length of segment $D E$.
As shown in Figure 6, draw $D F \perp A B$ at $F$.
From the given conditions, we easily know
$$
\begin{array}{c}
D F=\frac{\sqrt{3}}{2}, \\
A F=\frac{1}{2}, A E=\frac{2}{3}, \\
F E=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}, \\
D E=\sqrt{\left(\frac{1}{6}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\frac{\sqrt{7}}{3} .
\end{array}
$$

Therefore, the minimum value of $P E+P B$ is $\frac{\sqrt{7}}{3}$.