You do not need to reduce $ATM$ to $L$. As it hints, Let $L$ is recognized by a TM $M_L$, and note that $L = L(M_L)$ ($L(M_L)$ means the language recognized by $M_L$). Then you have the following contradictions $$\langle M_L \rangle \in L \implies \langle M_L \rangle \notin L(M_L) \implies \langle M_L \rangle \notin L$$ and $$\langle M_L \rangle \notin L \implies \langle M_L \rangle \in L(M_L) \implies \langle M_L \rangle \in L$$ This completes the proof.

You do not need to reduce $ATM$ to $L$. As it hints, Let $L$ is recognized by a TM $M_L$, and note that $L = L(M_L)$ ($L(M_L)$ denotes the language recognized by $M_L$). Then you have the following contradictions $$\langle M_L \rangle \in L \implies \langle M_L \rangle \notin L(M_L) \implies \langle M_L \rangle \notin L$$ and $$\langle M_L \rangle \notin L \implies \langle M_L \rangle \in L(M_L) \implies \langle M_L \rangle \in L$$ This completes the proof.