8. A.

Let the radius of the circumscribed sphere be $r$. Then
$$
(1-r)^{2}+\left(\frac{\sqrt{2}}{2}\right)^{2}=r^{2} \Rightarrow r=\frac{3}{4} \text {. }
$$

Thus, the surface area of the sphere is $4 \pi r^{2}=4 \pi\left(\frac{3}{4}\right)^{2}=\frac{9 \pi}{4}$.
8. Figure 2 shows the three views of a triangular pyramid, where the front view and side view are congruent isosceles triangles. Then the surface area of the circumscribed sphere of this triangular pyramid is ( ).
(A) $\frac{9 \pi}{4}$
(B) $9 \pi$
(C) $4 \pi$
(D) $\pi$