Proved as follows:

Let M(2, t), because a (-2, 0), then the equation of straight line AM is: y = (t/4) (x+2), what is the relationship between this straight line and ellipse X? /4+y？ /2= 1 simultaneous, eliminate y, and get: (t? +8)x？ +4t？ x+4t？ -32 = 0, so the equation has a root X 1 =-2, which is the abscissa of point A, so we use x 1=-2 = (4t? -32)/(t？ +8)，get: x2 = ( 16-2t？ )/(t？ +8), this is the abscissa of point P, and the coordinate of point P is (16-2t? )/(t？ +8)，8t/(t？ +8), and b (2 2,0), q (x x,0). Since the circle with a diameter of PM passes through the intersection of AM and BP, the vector QM is perpendicular to the vector BP, that is, QM*BP=0, while QM = (2-x, t) and BP = (-4t? )/(t？ +8)，8t/(t？ +8)), substituted, get: (2-x) [(-4t? )/(t？ +8)]+8t？ /(t？ +8) = 0, which means 4t? X=0 holds for all t∈R constants, then: x=0, so the coordinate of point Q is Q (0 0,0).