Solution:

The answer is $\mathbf{(C)}$. Let $O$ and $r$ be respectively the center and the radius of the circle $\Gamma$. Observe that the trapezoid must necessarily be isosceles (because it is inscribable in a circle), hence the diagonals are equal and $AP = PB$ (so also $PC = PD$). Therefore the triangles $APB$ and $BPC$ are isosceles and right-angled: they have area respectively $\frac{BP^2}{2}$ and $\frac{PC^2}{2}$.

Moreover, since triangle $APB$ is isosceles and right-angled, we find $45^\circ = \widehat{PAB} = \widehat{CAB} = \frac{\widehat{COB}}{2}$: that is, $\widehat{COB} = 90^\circ$. By the Pythagorean theorem on triangles $BPC$ and $BOC$ we find that: $2r^2 = BC^2 = BP^2 + PC^2$.

The sum of the areas of triangles $APB$ and $BPC$ equals $\frac{BP^2 + PC^2}{2} = r^2$ and the ratio sought is $\frac{\pi r^2}{r^2} = \pi$.