9. In $\square A B C D$, $\overrightarrow{A B} \cdot \overrightarrow{B D}=0$, fold along $B D$ to form a right dihedral angle $A-B D-C$, and $4 A B^{2}+2 B D^{2}=1$. Then the surface area of the circumscribed sphere of the triangular prism $A-B C D$ is ( ).
(A) $\frac{\pi}{2}$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{48}$
(D) $\frac{\sqrt{2} \pi}{24}$
9. A.

In the tetrahedron $A-BCD$, $AB \perp$ plane $BCD$, $\triangle BCD$ is a right triangle, and it is extended to form a rectangular prism.
$$
\begin{array}{l}
\text { Also, } AC=2R \text {, then } \\
AC^{2}=AB^{2}+BD^{2}+CD^{2} \\
=2AB^{2}+BD^{2}=\frac{1}{2}=4R^{2} . \\
\Rightarrow R^{2}=\frac{1}{8} .
\end{array}
$$

Therefore, the required result is $4 \pi R^{2}=\frac{\pi}{2}$.