24. Let $\triangle A B C$ be an equilateral triangle with side length $s$, and $P$ be a point inside the triangle, such that $A P=1, B P=\sqrt{3}, C P=2$. Then the value of $s$ is ( ).
(A) $1+\sqrt{2}$
(B) $\sqrt{7}$
(C) $\frac{8}{3}$
(D) $\sqrt{5+\sqrt{5}}$
(E) $2 \sqrt{2}$
24. B.

As shown in Figure 9, $\triangle A B P$ is rotated counterclockwise by $60^{\circ}$ around point $A$, resulting in $\triangle A C P^{\prime}$.
Thus, $\triangle A P P^{\prime}$ is an equilateral triangle.
In $\triangle P P^{\prime} C$, we have
$$
P^{\prime} C=\sqrt{3}, P P^{\prime}=1, P C=2 \text {, }
$$

which implies $\angle P P^{\prime} C=90^{\circ}, \angle A P^{\prime} C=150^{\circ}$.
Therefore, $s^{2}=A C^{2}$
$=A P^{\prime 2}+P^{\prime} C^{2}-2 A P^{\prime} \cdot P^{\prime} C \cos 150^{\circ}=7$
$\Rightarrow s=\sqrt{7}$.