1. We need to identify which compound gives oxygen gas, $O_2$, on hydrolysis with water.

2. Let us check each option.

Option A: $Pb_3O_4$

$Pb_3O_4$ is red lead, a mixed oxide of lead. It does not liberate $O_2$ on hydrolysis with water.

So, A is incorrect.

Option B: $KO_2$

$KO_2$ is potassium superoxide. Superoxides react with water as:

$4KO_2 + 2H_2O \rightarrow 4KOH + 3O_2$

Thus, hydrolysis of $KO_2$ liberates oxygen gas.

So, B is correct.

Option C: $Na_2O_2$

$Na_2O_2$ is sodium peroxide. On hydrolysis:

$Na_2O_2 + 2H_2O \rightarrow 2NaOH + H_2O_2$

It forms hydrogen peroxide, not directly $O_2$ on hydrolysis.

So, C is incorrect.

Option D: $Li_2O_2$

Similarly, lithium peroxide reacts as:

$Li_2O_2 + 2H_2O \rightarrow 2LiOH + H_2O_2$

Again, it gives $H_2O_2$, not directly $O_2$.

So, D is incorrect.

3. Therefore, the only compound that liberates $O_2$ upon hydrolysis is:

$\boxed{KO_2}$

4. Comparing with the stored correct answer: B.

My derived answer matches the stored answer.