The value of $$K_c$$ is 64 at 800 K for the reaction
$$\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$$
The value of $$K_c$$ for the following reaction is:
$$\text{NH}_3(g) \to \frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g)$$

We are given that, at 800 K, the equilibrium constant for the reaction

$$\text{N}_2(g)+3\text{H}_2(g)\;\longrightarrow\;2\text{NH}_3(g)$$

is $$K_c = 64$$.

First, recall two useful facts about equilibrium constants:

1. If a chemical equation is reversed, the new equilibrium constant is the reciprocal of the original: $$K_{\text{rev}} = \dfrac{1}{K_{\text{fwd}}}.$$ 2. If every coefficient in an equation is multiplied by a factor $$n$$, the new equilibrium constant is the old one raised to the power $$n$$: $$K_{\text{new}} = (K_{\text{old}})^{\,n}.$$

Now we convert the given reaction to the required one step by step.

Step 1 — Reverse the reaction. Reversing

$$\text{N}_2 + 3\text{H}_2 \longrightarrow 2\text{NH}_3$$

gives

$$2\text{NH}_3 \longrightarrow \text{N}_2 + 3\text{H}_2.$$

Using the first rule, the equilibrium constant for this reversed reaction is

$$K_{\,\text{rev}} = \frac{1}{64}.$$

Step 2 — Halve all stoichiometric coefficients. The required equation is

$$\text{NH}_3 \longrightarrow \tfrac12\text{N}_2 + \tfrac32\text{H}_2,$$

which is obtained by multiplying every coefficient in the last equation by the factor $$\tfrac12$$.

Applying the second rule with $$n = \tfrac12$$, we get

$$K_{\text{required}} = \left(K_{\,\text{rev}}\right)^{1/2} = \left(\frac{1}{64}\right)^{1/2}.$$

We now take the square root:

$$\left(\frac{1}{64}\right)^{1/2} = \frac{1}{\sqrt{64}} = \frac{1}{8}.$$

Hence, the correct answer is Option D.