1. For gaseous equilibria, the relation between $K_p$ and $K_c$ is

$K_p = K_c (RT)^{\Delta n_g}$

where $\Delta n_g$ is:

$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$

2. We need the equilibrium for which $K_p \ne K_c$.

This happens when:

$\Delta n_g \ne 0$

If $\Delta n_g = 0$, then $K_p = K_c$.

3. Check each option.

Option A

$2NO(g) \rightleftharpoons N_2(g) + O_2(g)$

Gaseous moles on reactant side $= 2$

Gaseous moles on product side $= 1+1=2$

$\Delta n_g = 2-2=0$

So,

$K_p = K_c$

Hence, A is not correct.

Option B

$2C(s) + O_2(g) \rightleftharpoons 2CO(g)$

Only gases are counted; solid carbon is ignored.

Gaseous moles on reactant side $= 1$

Gaseous moles on product side $= 2$

$\Delta n_g = 2-1=1$

So,

$K_p = K_c(RT)^1$

Thus,

$K_p \ne K_c$

Hence, B is correct.

Option C

$2HI(g) \rightleftharpoons H_2(g) + I_2(g)$

Gaseous moles on reactant side $= 2$

Gaseous moles on product side $= 1+1=2$

$\Delta n_g = 2-2=0$

So,

$K_p = K_c$

Hence, C is not correct.

Option D

$NO_2(g) + SO_2(g) \rightleftharpoons NO(g) + SO_3(g)$

Gaseous moles on reactant side $= 1+1=2$

Gaseous moles on product side $= 1+1=2$

$\Delta n_g = 2-2=0$

So,

$K_p = K_c$

Hence, D is not correct.

4. Therefore, the only equilibrium for which $K_p \ne K_c$ is:

$\boxed{\text{B}}$