Complexity of variation of partition problem
Is it in NP? Yes - any restricted version of an NP problem is also in NP.

Is it still NP-hard? Yes. 2-partition (without the constraint that both subsets should be equal cardinality) is NP-hard. Let's call your form of the problem "equal cardinalty 2-partition". To reduce from regular 2-partition to equal cardinalty 2-partition, we can pad the 2-partition problem by adding lots of extra 0's to the set, so that the 0's can be used to balance out any valid 2-partition solution to a solution that is an equal cardinality 2-partition.

You might object to the use of 0's in the input, and define "positive balanced cardinality 2-partition" where the input must be positive (so 0's are not allowed). But I can reduce any balanced cardinalty 2-partition problem to positive balanced cardinalty 2-partition by incrementing all numbers in the input by 1. This does not change the solution because of the equal cardinality constraint (each part of the partition gets increased by equally much).

Now you might still object that this reduction requires that you allow duplicate numbers in the input set. Even if you disallow duplicates in the input it remains NP-hard, and you can show so by a reduction from Positive 1-IN-3 SAT.

Is it in P? Answering this question would settle the P vs. NP problem so you're not very likely to get an answer any time soon.