Preparation of potassium permanganate from $$MnO_2$$  involves two step process in which the 1st step is a reaction  with $$KOH$$ and  $$KNO_3$$ to produce:

The preparation of potassium permanganate ($$KMnO_4$$) from $$MnO_2$$ involves two steps. First, $$MnO_2$$ is fused with $$KOH$$ in the presence of an oxidizing agent like $$KNO_3$$ (or atmospheric oxygen):

$$2MnO_2 + 4KOH + O_2 \to 2K_2MnO_4 + 2H_2O$$

In this step, $$Mn$$ is oxidized from +4 to +6 oxidation state, forming potassium manganate ($$K_2MnO_4$$), which is a green compound.

Then, $$K_2MnO_4$$ is oxidized to $$KMnO_4$$ (Mn goes from +6 to +7) by electrolytic oxidation or treatment with chlorine or ozone.

The product of the first step is $$K_2MnO_4$$.

The correct answer is Option D: $$K_2MnO_4$$.