The expression after parabolic translation is y=-(x+ 1)(x-3)=-x2+2x+3, that is, y=-x2+2x+3.
∫y =-x2+2x+3 =-(x- 1)2+4，
∴ The coordinate of vertex D is (1, 4);
(2)ACB equals ABD for the following reasons:
As shown ∫y =-x2+2x+3,
∴ When point x=0 and y=3, the coordinate of point C is (0,3).
∫b(3，0)，∠ BOC = 90，
∴OB=OC,∠OBC=∠OCB=45。
In △BCD, ∫BC2 = 32+32 = 18, CD2= 12+ 12=2, BD2=22+42=20,
∴BC2+CD2=BD2，
∴∠BCD=90，
∴tan∠cbd=cdbc=2 18= 13，
∫In△AOC，∠ AOC = 90，
∴tan∠ACO=OAOC= 13，
∴tan∠ACO=tan∠CBD，
∴∠ACO=∠CBD，
∴∠ACO+∠OCB=∠CBD+∠OBC，
That is, ∠ ACB = ∠ Abd;
(3)∵ Point P is on the axis of symmetry of translation parabola, and the axis of symmetry of y=-x2+2x+3 is x= 1.
∴ The coordinate of point P is (1, n).
∵△ABC is an acute triangle,
When △CDP is similar to △ABC, △CDP is also an acute triangle.
∴ n < 4, that is, point P can only be below point D,
∠∠CDP =∠ABC = 45，
∴D and B are corresponding points, which are divided into two situations:
① If △CDP∽△ABC, then CDAB=DPBC.
That is 24=4? N32, the solution is n=52,
∴ The coordinate of point P is (1, 52);
② If △CDP∽△CBA, then CDCB=DPAB.
That is 232=4? N4, the solution is n=83,
∴ The coordinate of point P is (1, 83).
To sum up, the coordinates of point P are (1, 52) or (1, 83).