Solution
As shown in the figure, let the midpoint of $BC$ be $F$, and connect $AF, DF$. Then $\triangle A_{1}PA \sim \triangle ADF \Rightarrow \frac{A_{1}P}{AD}=\frac{A_{1}A}{AF} \Rightarrow AD=\frac{A_{1}P \cdot AF}{A_{1}A}$. Since $A_{1}P \cdot AF=\frac{2}{3}AF^2=\frac{2}{3} \cdot\left(\frac{\sqrt{3}}{2}\right)^2=\frac{1}{2} \Rightarrow AD=\frac{1}{4}$, thus $V_{D-ABC}=\frac{1}{3} \cdot \frac{\sqrt{3}}{4} \cdot \frac{1}{4}=\frac{\sqrt{3}}{48}$. Therefore, the answer is $A$.
2. Given point $P$ is the center of the upper base $A_{1} B_{1} C_{1}$ of a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, construct plane $B C D \perp A P$, intersecting edge $A A_{1}$ at $D$. If $A A_{1}=2 A B=2$, then the volume of the tetrahedron $D-A B C$ is
A. $\frac{\sqrt{3}}{48}$
B. $\frac{\sqrt{3}}{24}$
C. $\frac{\sqrt{3}}{16}$
D. $\frac{\sqrt{3}}{12}$