Alas, this kind of topic is far from textbooks and college entrance examination. You don't have to look at the college entrance examination. If you want to finish this question in 15 minutes, you will get 17 points. It is not easy! I'm already tired.

Because PA is perpendicular to the right-angled ABCD, and J-angled ABCD happens to be the equatorial circle of the ball. So, PA is the tangent of the ball and the circle. As shown in the figure. Note that the side PB of the isosceles right triangle PBD is also the tangent of the circle, PB=BD, M is on the circle, so OM is the center line of the triangle. OD=OM, OM//PB, om = half OM=PB.

MN//CD=AB，？ MNBA is a cross section.

M is the midpoint of PD. With the above analysis, we can prove it. Again, don't look at him.