Starting from a text book description of the cross-coupled pair, when one of the input goes high by delta_v amount and another input goes low by delta_v amount, the tail node voltage does not change and the (small-signal) current through one transistor is given by gm(delta_v) and the through the other transistor by gm(-delta_v). The small signal current flows in a circular loop.
Now when Vx=Vy in your case, both the transistors need to carry the same current. The drop in Vp is because of the Vgs requirement of M1 and M2 to carry Itail/2. The value of Vp you see is roughly the DC voltage you would see if you had tied both your inputs to the same voltage (=Vx=Vy). But as Vx starts moving away further and further higher and Vy the other way, because of the square law nature of the device, Vp needs to go back up.