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.