2. Given a regular triangular prism $A B C-A_{1} B_{1} C_{1}$ with a height of 2 and a base edge length of 1, the center of the upper base $\triangle A_{1} B_{1} C_{1}$ is $P$. A plane $B C D \perp A P$ is made through the lower base edge $B C$, intersecting the edge $A A_{1}$ at point $D$. Then the area of the section $\triangle B C D$ is $\qquad$
2. $\frac{\sqrt{13}}{8}$.

As shown in Figure 1, let the plane $A A_{1} P$ intersect the edges $B C$ and $B_{1} C_{1}$ at points $E$ and $E_{1}$, respectively.
As shown in Figure 2, in the rectangle $A E E_{1} A_{1}$, it is easy to know that $A A_{1}=2, A E=A_{1} E_{1}=\frac{\sqrt{3}}{2}, A_{1} P=\frac{\sqrt{3}}{3}$. Let $\angle A_{1} A P=\angle A E D=\theta$.
Then $\tan \theta=\frac{A_{1} P}{A A_{1}}=\frac{\sqrt{3}}{6}$.
Thus, $\cos \theta=\sqrt{\frac{12}{13}} \Rightarrow D E=\frac{A E}{\cos \theta}=\frac{\sqrt{13}}{4}$
$\Rightarrow S_{\triangle B C D}=\frac{1}{2} D E \cdot B C=\frac{\sqrt{13}}{8}$.