In KO$$_2$$, the nature of oxygen species and the oxidation state of oxygen atom are, respectively:

We start with the compound $$\mathrm{KO_2}$$. Potassium belongs to the alkali-metal group and in all its stable compounds it carries an oxidation number of $$+1$$. We write this fact explicitly:

$$\text{Oxidation number of K}=+1$$

Let the oxidation number of each oxygen atom in $$\mathrm{KO_2}$$ be $$x$$. There are two oxygen atoms, so the total contribution of oxygen to the overall charge is $$2x$$.

The compound $$\mathrm{KO_2}$$ is electrically neutral, therefore the algebraic sum of the oxidation numbers of all the atoms present must be equal to the net charge, which is zero. Using the general rule

$$\sum (\text{oxidation numbers})=\text{overall charge},$$

we can write

$$+1 + 2x = 0.$$

Now we solve this simple linear equation step by step.

Subtract $$1$$ from both sides:

$$2x = -1.$$

Divide both sides by $$2$$ to isolate $$x$$:

$$x = \frac{-1}{2}.$$

So, each oxygen atom in $$\mathrm{KO_2}$$ has an oxidation state of $$- \dfrac{1}{2}.$$

Next, we classify the type of oxygen species. The three common dioxygen anions are:

1. $$\mathrm{O^{2-}}$$  — oxide ion (oxidation state of each O = $$-2$$)
2. $$\mathrm{O_2^{2-}}$$  — peroxide ion (oxidation state of each O = $$-1$$)
3. $$\mathrm{O_2^{-}}$$  — superoxide ion (oxidation state of each O = $$-\dfrac12$$)

Because we have just calculated an oxidation state of $$-\dfrac12$$ for each oxygen atom, the dioxygen fragment must be the superoxide ion $$\mathrm{O_2^{-}}$$.

Therefore, in $$\mathrm{KO_2}$$ the oxygen species is a superoxide, and the oxidation state of each oxygen atom is $$-\dfrac12$$.

Hence, the correct answer is Option B.