I don't know if I'm missing something..Let $\ell(X)=\text{label of set }X$. Consider the set $G$ of all pairs $(T, \ell(T))$ where $ T\in\mathcal{P} $ such that $\ell(T)\notin T$. Clearly, $|\mathcal{P}|={100\choose 49}\ge |G|$. Suppose for the sake of contradiction that no such $M$ exists. It's rather easy to see that a $50$-element set $M$ has not the required property iff we can find $(T, \ell(T))\in G$ so that $M=T\cup\ell(T)$. Each element of $G$ can form at most one such set $M$. But ${100\choose 50}>{100\choose 49}\ge |G|$, contradiction.