The coordinate is (m,-2m+b); Point D and point B are symmetrical about the origin, so the coordinate of point D is (-m, 2m-b);

The slope of the straight line where BD is located is kbd = [(2m-b)-(-2m+b)]/(-m-m) = (4m-2b)/(-2m) = (b-2m)/m.

For the vertical line whose origin is BD, its slope is =-m/(b-2m), its equation is y=-[m/(b-2m)]x, and let x=m, that is, the vertical position of point P.

Mark y=-m? /(b-2m), that is, the coordinates of point P are (m, -m? /(b-2m))..........( 1)

At this time, P is on the vertical line of BD, so there must be PB=PD, that is, △PBD is an isosceles triangle.

Because AB⊥BC, KAB=-2 and the coordinate of point C is (0, -b), KBC = [(-2m+b)-(-b)]/m = (-2m+2b)/m =1/2.

-4m+4b=m, 5m=4b, ∴m=4b/5. When the formula (1) is substituted, the coordinate of point P is (4b/5, 16b/ 15).