I am familiar with the mechanism of proof by contradiction: we want to prove $P$, so we assume $¬P$ and prove that this is false; hence $P$ must be true.
I have the following devil's advocate question, which might seem to be more philosophy than mathematics, but I would prefer answers from a mathematician's point of view:
When we prove that $¬P$ is "false", what we are really showing is that it is inconsistent with our underlying set of axioms. Could there ever be a case were, for some $P$ and some set of axioms, $P$ and $¬P$ are both inconsistent with those axioms (or both consistent, for that matter)?
The situation you ask about, where $P$ is inconsistent with our axioms and $\neg P$ is also inconsistent with our axioms, would mean that the axioms themselves are inconsistent. Specifically, the inconsistency of $P$ with the axioms would mean that $\neg P$ is provable from those axioms. If, in addition, $\neg P$ is inconsistent with the axioms, then the axioms themselves are inconsistent --- they imply $\neg P$ and then they contradict that. (I have phrased this answer so that it remains correct even if the underlying logic of the axiom system is intuitionistic rather than classical.)