What determines orbital energy in many-electron atoms?
The energy of an orbital depends on the effective nuclear charge $Z_{\text{eff}}$ felt by that orbital and on shielding/penetration effects.

For the $2s$ orbital in the given atoms:

• $\mathrm{H}$ has configuration $1s^1$.

  • It does not have a filled/occupied $2s$ orbital in the ground state, but the hydrogenic $2s$ level exists.
  • For hydrogen, orbital energy depends only on principal quantum number $n$, so $E_n=-\frac{13.6}{n^2}\,\text{eV}$ Hence, $E_{2s}(\mathrm{H})=E_{n=2}=-\frac{13.6}{4}=-3.4\,\text{eV}.$

• $\mathrm{Li}$: $1s^2 2s^1$

• $\mathrm{Na}$: $1s^2 2s^2 2p^6 3s^1$

• $\mathrm{K}$: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$

In multi-electron atoms, inner orbitals such as $2s$ become more stabilized (more negative in energy) as atomic number increases, because they experience a larger nuclear attraction.

Compare $2s$ orbital energy across $\mathrm{Li}$, $\mathrm{Na}$, and $\mathrm{K}$:

• As we go from $\mathrm{Li} \to \mathrm{Na} \to \mathrm{K}$, nuclear charge increases.
• The $2s$ orbital is a core orbital in $\mathrm{Na}$ and $\mathrm{K}$ and is held much more strongly.
• Therefore, its energy becomes progressively lower: $E_{2s}(\mathrm{K}) < E_{2s}(\mathrm{Na}) < E_{2s}(\mathrm{Li}).$

Compare with hydrogen:

• In hydrogen, $E_{2s}=-3.4\,\text{eV}$.
• In heavier atoms, the $2s$ orbital is much more tightly bound than this, so its energy is lower than in hydrogen as well.

Conclusion:
The lowest energy of the $2s$ orbital is in $\boxed{\mathrm{K}}.$

Option check:

• A: Li — not lowest
• B: H — not lowest
• C: Na — lower than Li, but not lowest
• D: K — correct