Note that the condition for $\Gamma$ in the problem is equivalent to the following condition: if $\min (\angle A P B, \angle B P C, \angle C P D, \angle D P A)>60^{\circ}$, then $P$ is contained in the interior of $\Gamma$. Let $X_{1}, X_{2}, X_{3}$, and $X_{4}$ be the four points in $A B C D$ such that $A B X_{1}, B C X_{2}, C D X_{3}$, and $D A X_{4}$ are all equilateral triangles. Now, let $\Omega_{1}, \Omega_{2}, \Omega_{3}$, and $\Omega_{4}$ be the respective circumcircles of these triangles, and let the centers of these circles be $O_{1}, O_{2}, O_{3}$, and $O_{4}$. Note that the set of points $P$ such that $\angle A P B, \angle B P C, \angle C P D, \angle D P A>60^{\circ}$ is the intersection of $\Omega_{1}, \Omega_{2}, \Omega_{3}$, and $\Omega_{4}$. We want to find the area of the minimum circle containing this intersection. Let $\Gamma_{1}$ and $\Gamma_{2}$ intersect at $B$ and $B^{\prime}$. Define $C^{\prime}, D^{\prime}$ and $A^{\prime}$ similarly. It is not hard to see that the circumcircle of square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is the desired circle. Now observe that $\angle A B^{\prime} D^{\prime}=\angle A B^{\prime} D=60^{\circ}$. Similarly, $\angle A D^{\prime} B^{\prime}=60^{\circ}$, so $A B^{\prime} D^{\prime}$ is equilateral. Its height is the distance from $A$ to $B^{\prime} D^{\prime}$, which is $\frac{1}{\sqrt{2}}$, so its side length is $\frac{\sqrt{6}}{3}$. This is also the diameter of the desired circle, so its area is $\frac{\pi}{4} \cdot \frac{6}{9}=\frac{\pi}{6}$.