Solution:

The answer is $(\mathbf{A})$. Let $O$ be the center of the cardboard circle, $O'$ the point where Priscilla places the point of the compass, $A$ and $B$ the intersections of the arc of a circle drawn by Priscilla with the edge of the circle. Then $A O' B$ is an isosceles right triangle (we can justify this by noting that the triangles $O O' A$ and $O O' B$, since they have sides of length $r, r$ and $\sqrt{2} r$, are both isosceles right triangles). Let $A_{A O' B}$ be the area of the circular sector $A O' B$, $A_{C}$ that of the cardboard circle and $A_{T}$ that of the triangle $A O' B$. Then the area of the crescent moon is equal to $A_{C}-A_{A O' B}-\frac{1}{2} A_{C}+A_{T}$, hence to $\pi r^{2}-\frac{1}{4} \pi(\sqrt{2} r)^{2}-\frac{1}{2} \pi r^{2}+\frac{1}{2}(\sqrt{2} r)^{2}$, that is, exactly $r^{2}$.