Answer is (B) NP-complete $\cap P = \phi$
Since, P $\neq$ NP, there is at least one problem in NP, which is harder than all P problems. Lets take the hardest such problem, say $X$. Since, P $\neq$ NP, $X \notin $ P .
Now, by definition, NP-complete problems are the hardest problems in NP and so $X$ problem is in NP-complete. And being in NP, $X$ can be reduced to all problems in NP-complete, making any other NP-complete problem as hard as $X$. So, since $X \notin $P, none of the other NP-complete problems also cannot be in P.