the locus of any point P which is equidistant from two intersecting lines is the angular bisector of the included angle between the two lines .

here,as P is equidistant from AB and BC,draw the bisector of angle ABC.name the line BD.now the point P lies somewhere on line BD.

Now P is also equidistant from from A and B.so,draw the perpendicular bisector of AB.extend the bisector enough to intersect BD.thus,the point where the perpendicular bisector of AB meets angle bisector BD,is the exact location of point P.

its because :that will form two triangles PIA and PIB ,which will be congruent by side-angle-side congruency,and thus PA=PB

hope that it is correct and helps!!!