This pic clears this question.
(a) As NP complete problem are hardest problem in NP class.Every NP complete problem is in NP, but vice versa is not true. Moreover NP Complete problem also should be satisfy NP hard property. So, all NP problem shouldnot reduced to NP Complete.
A fast solution to any NP-Complete problem would mean that as long as you can verify proposed solutions to a problem you would never need to search through a substantial fraction of the search space to find solutions; there would always be a faster way. Therefore most mathematician thought there is no solution faster than NP complete (Which cannot be a case for NP)
(b) If a problem is in NP Complete then it must be in NP . Means if X is a NP Complete problem , then X must be in NP. Here NP!=P , that is why $$NPComplete\cap P=\phi$$
(c) NP Hard=NP
What does NP-hard mean? A lot of times you can solve a problem by reducing it to a different problem. I can reduce Problem B to Problem A if, given a solution to Problem A, I can easily construct a solution to Problem B. (In this case, "easily" means "in polynomial time.")
Only NP complete problem polynomial time reducible to NP Hard problem. But all NP problem not in NPC. So, all NP problem cannot be reduced to NP Hard .http://gatecse.in/p-np-np-complete-np-hard/
(d) NP Complete problem is the set of problems, for which NP problems can be reduced to NP Complete.And we know . If NP problem can be solved quickly, then NP Complete problem also can solved quickly.And according to the diagram sometimes P=NP and sometimes P!=NP. NP complete is a subset of NP. So, P=NP Complete is not correct equation.