Ok i've written it with full proof if you are still unsure how to obtain the answer. This is the most basic way to probably do it, doesn't require too much thinking
i)
S BC = BG (given sides of square BCFG are equal
angle DBA = angle CBG =90
angle ABC + angle DBA = angle ABC + andgle CBG since Angle abc is common
angle abc = angle dba = angle dbc
angle abc + angle cbg = angle abg
ie angle dbc = angle abg
ab = pb (sides of square)
therefore triangle abg == triangle dbc (SAS)
ii)
angle abg = angle dcb (corresponding angle's in cong triangles)
therefore since the angles subtend by the arc in a cong triangle, pb are equal it is now known that P, C, B, G lie on the cercomprence of a circle thus making them cyclic.
iii)
angle cpg = angle cbg = 90 (angles subtended by the arc cg are euqal)
therefore AG meets dc at 90 degrees, they are perpendicular
iv)
angle cgf = angle cgb = 45 (line cutting a square diagionally bisects the angle)
note: if you dont like this reason you can prove the 2 triangles are congruant etc...
but that is silly and a waste of time..
therefore angle dbp = angle cgb = 45 (exterrior opposing angles in a cyclic quad are euqal)
therefor angle dpg - angle dpb = 45
that is: (90 - 45 = 45)
therefore angle dpb = angle bpg = 45
PB bisects angle dpg