7. As shown in Figure 1, in a regular tetrahedron $A B C D$ with edge length 1, $G$ is the centroid of $\triangle B C D$, and $M$ is the midpoint of line segment $A G$. Then the surface area of the circumscribed sphere of the tetrahedron $M$ $-B C D$ is ( ).
(A) $\pi$
(B) $\frac{3 \pi}{2}$
(C) $\frac{\sqrt{6} \pi}{4}$
(D) $\frac{\sqrt{6} \pi}{8}$
7. B.

Since $G$ is the centroid of the equilateral $\triangle B C D$, we have $A G \perp$ plane $B C D \Rightarrow A G \perp B G$.
In the right $\triangle A G B$,
$A B=1, B G=\frac{2}{3} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}$
$\Rightarrow A G=\sqrt{A B^{2}-B G^{2}}=\frac{\sqrt{6}}{3}$
$\Rightarrow M G=\frac{1}{2} A G=\frac{\sqrt{6}}{6}$.
In the right $\triangle M G B$,
$M B=\sqrt{M G^{2}+B G^{2}}=\frac{\sqrt{2}}{2}$
$\Rightarrow M C=M B=\frac{\sqrt{2}}{2}$.
Thus, $M B^{2}+M C^{2}=1^{2}=B C^{2} \Rightarrow M B \perp M C$.
Similarly, $M C \perp M D, M D \perp M B$.
Therefore, the diameter of the circumscribed sphere of the tetrahedron $M-B C D$ is equal to the body diagonal of a cube with edges $M B, M C, M D$.
Let the radius of the circumscribed sphere of the tetrahedron $M-B C D$ be $R$.
Then $2 R=\sqrt{3} M B=\frac{\sqrt{6}}{2}$.
Thus, the surface area of the circumscribed sphere $S=4 \pi R^{2}=\frac{3 \pi}{2}$.