2. Given that point $E$ is the midpoint of side $B C$ of rhombus $A B C D$, $\angle A B C=30^{\circ}$, and $P$ is a point on diagonal $B D$ such that $P C + P E = \sqrt{13}$. Then the maximum area of rhombus $A B C D$ is $\qquad$
$2.10+4 \sqrt{3}$.
As shown in Figure 3, connect $A C$, $A E$, and $A P$, and draw $A F \perp B C$ at point $F$.
From the problem, we have $A B=B C$, and $B D$ is the perpendicular bisector of $A C$. Therefore, $P A=P C$.
Also, $A E \leqslant P A+P E=P C+P E=\sqrt{13}$, so the maximum value of $A E$ is $\sqrt{13}$.
Let $A B=2 a$.
In the right triangle $\triangle A B F$, since $\angle A B C=30^{\circ}$, it is easy to know that $A F=a$ and $B F=\sqrt{3} a$.
Also, $B E=\frac{1}{2} B C=a$, thus,
$$
E F=B F-B E=(\sqrt{3}-1) a \text{. }
$$

In the right triangle $\triangle A E F$, by $A F^{2}+E F^{2}=A E^{2}$, we get $a^{2}+[(\sqrt{3}-1) a]^{2}=A E^{2}$.
Then $a^{2}=\frac{A E^{2}}{5-2 \sqrt{3}}=\frac{5+2 \sqrt{3}}{13} A E^{2} \leqslant 5+2 \sqrt{3}$.
Therefore, the area of rhombus $A B C D$ is
$$
B C \cdot A F=2 a \cdot a=2 a^{2} \leqslant 10+4 \sqrt{3} \text{. }
$$