For the reaction
$$\text{Fe}_2\text{N}(s) + \frac{3}{2}\text{H}_2(g) \rightleftharpoons 2\text{Fe}(s) + \text{NH}_3(g)$$

We begin by writing the equilibrium constant expressions in terms of concentration and pressure. Because pure solids do not appear in equilibrium expressions, only the gaseous species contribute.

So, for the reaction
$$\text{Fe}_2\text{N}(s) + \frac32\text{H}_2(g) \rightleftharpoons 2\text{Fe}(s) + \text{NH}_3(g)$$ we have

$$K_c \;=\;\frac{[\text{NH}_3]}{[\text{H}_2]^{3/2}}$$

and

$$K_p \;=\;\frac{P_{\text{NH}_3}}{P_{\text{H}_2}^{\,3/2}}.$$

Next we recall the general relation that connects the two equilibrium constants:

$$K_p \;=\;K_c\,(RT)^{\Delta n_{\text{gas}}},$$

where $$\Delta n_{\text{gas}}$$ is the difference between the total number of moles of gaseous products and gaseous reactants.

For our reaction, gaseous products contribute $$1$$ mole (only $$\text{NH}_3(g)$$) and gaseous reactants contribute $$\tfrac32$$ moles (only $$\text{H}_2(g)$$). Hence

$$\Delta n_{\text{gas}} \;=\;1-\frac32 \;=\;-\frac12.$$

Substituting this value into the formula gives

$$K_p \;=\;K_c\,(RT)^{-\,\frac12}.$$

Now we solve for $$K_c$$ by multiplying both sides by $$(RT)^{+\,\frac12}:$$

$$K_c \;=\;K_p\,(RT)^{+\,\frac12}.$$ So we can write simply

$$K_c \;=\;K_p(RT)^{1/2}.$$

Hence, the correct answer is Option C.