4. In rhombus $A B C D$, $\angle A B C=60^{\circ}, A B=$ $1, E$ is the midpoint of side $B C$. Then the minimum value of the sum of the distances from a moving point $P$ on diagonal $B D$ to points $E$ and $C$ is ( ).
(A) $\frac{\sqrt{3}}{4}$
(B) $\frac{\sqrt{3}}{3}$
(C) $\frac{\sqrt{3}}{2}$
(D) $\sqrt{3}$
4.C.

As shown in Figure 3, construct the symmetric point $E_{1}$ of $E$ with respect to $BD$. Since $BD$ is the axis of symmetry of the rhombus $ABCD$, $E_{1}$ is the midpoint of side $AB$. Connect $CE_{1}$ and let it intersect $BD$ at point $P_{1}$.

Since $PE + PC = PE_{1} + PC \geqslant CE_{1}$, the equality holds when point $P$ coincides with $P_{1}$.
Given $\angle ABC = 60^{\circ}$, we know that $\triangle ABC$ is an equilateral triangle.
Therefore, $CE_{1} = \sqrt{1^{2} - \left(\frac{1}{2}\right)^{2}} = \frac{\sqrt{3}}{2}$.