Which of the following liberates O2 upon hydrolysis?

Hydrolysis means reaction with water at ordinary temperature.

Case A: $$Pb_3O_4$$ (red lead) is a mixed oxide of lead. It is practically insoluble in cold water and does not react with water to give any gaseous product. Hence no $$O_2$$ is formed.

Case B: $$KO_2$$ is a super-oxide. Super-oxides react with water according to the general relation
$$4MO_2 + 2H_2O \rightarrow 4MOH + 3O_2$$
For potassium super-oxide
$$4KO_2 + 2H_2O \rightarrow 4KOH + 3O_2$$
Thus $$O_2$$ is liberated directly during hydrolysis.

Case C: $$Na_2O_2$$ is a peroxide. Its hydrolysis is
$$Na_2O_2 + 2H_2O \rightarrow 2NaOH + H_2O_2$$
No molecular oxygen appears in this step; only hydrogen peroxide is produced.

Case D: $$Li_2O_2$$ (another peroxide) behaves similarly:
$$Li_2O_2 + 2H_2O \rightarrow 2LiOH + H_2O_2$$
Again, no $$O_2$$ is released during the hydrolysis itself.

Only the super-oxide $$KO_2$$ gives molecular oxygen directly on coming in contact with water.

Option B which is: $$KO_{2}$$