Make a circle with the midpoint of AC (set to D) as the center and AC/2 as the radius. (pictured)

Because PA 2+PC 2 = AC 2, the angle APC=90 degrees.

P must be on the circle.

When b, p and d are three points and one line, PB has a minimum value.

(DB=PB+PD, PD=AC/2 is a constant, and BD is the shortest when it is a line segment, and PB has a minimum value at this time)

At this time: CD=√3, BC=3? Angle DCB=90 degrees

The angle DBC is 30 degrees and the angle BDC is 60 degrees.

So: PDC is an equilateral triangle. The angle PCA is 60 degrees.

PC=√3，PA=3

The area of triangle ACP =(3*√3)/2= 1.5√3.