∵eb=pb=2、eb⊥pb,∴∠bep=45、pe=√2pb=2√2 .

Abcd is a square, ∴ AB = CB, ∠ ABC = 90.

∴∠pba=∠abc-∠pbc=90-∠PBC =∠pbe-∠PBC =∠ebc .

From AB = CB, Pb = EB, ∠ PBA = ∠ EBC, we get: △ PBA △ EBC, ∴ PA = EC = 1.

∵ec= 1、pe=2√2、pc=3,∴ec^2+pe^2=pc^2,∴∠cep=90，

∴∠bec=∠cep+∠bep=90 +45 .

From the cosine theorem, there are:

bc^2=eb^2+ec^2-2eb×eccos∠bec=4+ 1-2×2× 1×cos(90+45)= 5+2×√2 .

∴s(abcd)=bc^2=5+2×√2。