Determining range with Cauchy inequality.

∫A, B, C and D are all positive numbers, and a+b+c+d= 1.

∴(3a+ 1)+(3b+ 1)+(3c+ 1)+(3d+ 1)=7

∴ From Cauchy inequality:

28 = 4×7 =( 1+ 1+ 1+ 1)×[(3a+ 1)+(3 b+ 1)+(3c+ 1)+(3d+ 1)]

⊙[√(3a+ 1)+√(3b+ 1)+√(3c+ 1)+√(3d+ 1)]？

That is, there is always p≦2√7. The equal sign can only be obtained when a=b=c=d= 1/4.

This shows that it is possible that p = 2 √ 7 > 5.

But when A = 1/2 and B = C = D = 1/6, P = (√ 10+3 √ 6)/2 < 5.

The relationship between p and 5 is uncertain,

Choose D.