The main oxides formed on combustion of Li, Na and K in excess of air are respectively:

We begin by recalling a general qualitative rule of the periodic table: when alkali-metal cations $$M^+$$ are heated in excess oxygen, the nature of the anion of oxygen (oxide $$O^{2-}$$, peroxide $$O_2^{2-}$$ or superoxide $$O_2^-$$) that is stabilised depends strongly on the ionic radius of $$M^+$$.

The rule is:

Smaller $$M^+$$  – greater lattice energy with the small $$O^{2-}$$ ion  ⇒  normal oxide is favoured.

Medium $$M^+$$  –  lattice energy no longer sufficient to stabilise very highly charged $$O^{2-}$$, but still enough for $$O_2^{2-}$$  ⇒  peroxide is favoured.

Larger $$M^+$$  –  still lower lattice energy; even $$O_2^{2-}$$ is not comfortably stabilised, but the larger, less-charged $$O_2^-$$ matches well  ⇒  superoxide is favoured.

Let us apply this step by step to lithium, sodium and potassium.

1. Lithium has the smallest cation among the three. Because of its small radius, the lattice energy of $$Li_2O$$ (which contains $$O^{2-}$$) is very high and easily compensates for the high charge density of $$O^{2-}$$. Thus, in excess oxygen, the principal product is the normal oxide

$$2\,Li \;+\; \dfrac12\,O_2 \;\rightarrow\; Li_2O.$$

2. Sodium is larger than lithium. The lattice energy released on forming $$Na_2O$$ is now insufficient to offset the very high negative charge density of $$O^{2-}$$. However, the peroxide anion $$O_2^{2-}$$ has a lower charge density and a larger size, giving a better radius match with $$Na^+$$. Therefore sodium burns mainly to the peroxide:

$$2\,Na \;+\; O_2 \;\rightarrow\; Na_2O_2.$$

3. Potassium is still larger. For the even bigger $$K^+$$ the lattice energy becomes still smaller, so even the peroxide anion is not the best match. The superoxide anion $$O_2^-$$, which carries only a single negative charge and has the largest size among the three oxygen species, fits $$K^+$$ best. Consequently potassium gives the superoxide:

$$K \;+\; O_2 \;\rightarrow\; KO_2.$$

Collecting the three main oxides obtained in excess air, we have

$$Li \longrightarrow Li_2O, \quad Na \longrightarrow Na_2O_2, \quad K \longrightarrow KO_2.$$

Comparing this sequence with the options provided, we see that Option B lists exactly $$Li_2O$$, $$Na_2O_2$$ and $$KO_2$$ in that order.

Hence, the correct answer is Option B.