And I found out that proving P Vs NP is an NP problem.
I think the two different meanings of the word "problem" are mixed here.
When we discuss P and NP, they are the class of decision problems, which are essentially maps from the set of bits sequence to boolean values.
On the other hand, "NP =/= P" is a long standing mathematical problem (general meaning,) but "NP =/= P" itself is a single proposition, and usually it's not considered as decision problem.
If you convert "NP =/= P" to a decision problem as the map from single point to true/false (this is completely valid formalization), its complexity is constant time (algorithm can simply return true or false, though we don't know which one is correct,) so it's in P and NP.
if somebody proves that, then won't they be contradicting the result itself
Proving "NP =/= P" means we know the correct return value in the algorithm above, and that doesn't contradict with anything. The result simply claims there is a NP problem that is not P. The decision problem "NP =/= P" is just an instance within class P. Both facts can hold simultaneously.