Good question. Classical reasoning in natural deduction may be nonobvious. I'll describe it informally at first. I believe you can use double-negation elimination rule (DNE), which derives Q from ~~Q.

Assume ~(P \/ ~P). Then assuming P leads to a contradiction, so ~P. However, this also leads to a contradiction. Thus, ~~(P \/ ~P), from where you get P \/ ~P by DNE.