smc

Using the tangent-tangent theorem, $PA=AB=P{A}^{\prime }=A^{\prime }{B}^{\prime }=4$. We can then drop perpendiculars from the centers of the circles to the points of tangency and use similar triangles. Let us let the center of the smaller circle be point $S$ and the center of the larger circle be point $L$. If we let the radius of the larger circle be $x$ and the radius of the smaller circle be $y$, we can see that, using similar triangle, $x=2y$. In addition, the total hypotenuse of the larger right triangles equals $2(x+y)$ since half of it is $x+y$, so $y^2+4^2=(3y{)}^2$. If we simplify, we get $y^2+16=9y^2$, so $8y^2=16$, so $y=\sqrt{2}$. This means that the smaller circle has area $2\pi$, which is answer choice $\boxed{2\pi }$.