$$H_2O_2$$ acts as a reducing agent in

$$H_2O_2$$ acts as a reducing agent when it is oxidized, i.e., when the oxygen in $$H_2O_2$$ (oxidation state $$-1$$) is converted to a higher oxidation state.

Examining the reactions:

Option 1: $$2NaOCl + H_2O_2 \rightarrow 2NaCl + H_2O + O_2$$

Here, oxygen in $$H_2O_2$$ (oxidation state $$-1$$) is converted to $$O_2$$ (oxidation state $$0$$). The oxygen is oxidized, so $$H_2O_2$$ acts as a reducing agent.

In the other options, $$H_2O_2$$ acts as an oxidizing agent (oxygen goes from $$-1$$ to $$-2$$).