Solution: (2) See the figure below, let PF⊥AD be in F, then CF, ME⊥GF be in E, CG⊥AD be in G, then PG; It can be seen that ABCD is a cube AB = AP = PGAF=FG=AB/2, and △PGC is an isosceles right triangle; △PAG is an equilateral triangle; Then:? PC = PB =√2AB； Let PH⊥BC be H, and pH = √ (PC 2+HC 2) = √ (2+1/4) AB = 3ab/2; ?

Because PM=PC/3, MC = (1-1/3) PC = 2pc/3; s△mbc=( 1/2)bc*(2/3)pc=ab^2/2=2√7/3；

ab^2=4√7/3； AB = √( 4√7/3)= 2 √(√7/3)； ? pf=√(pg^2-fg^2)=√3ab/2；

vp-abcd=( 1/3)[(bc+ad)*ab/2]*pf=( 1/3)*( 1/2)*(3ab^3)*(√3/2)=(√3/4)*(4√7/3)*(2√(√7/3)

=2√(7√7)/3。