The question statement appears to be missing the actual two reversible chemical reactions, so the equilibrium constants $K_1$ and $K_2$ are not defined.

Without the reactions, we cannot derive any valid relation between $K_1$ and $K_2$.

However, we can still inspect the options for likely intended forms.

1. Standard rules for equilibrium constants

If a reaction is: $aA + bB \rightleftharpoons cC + dD$ with equilibrium constant $K$, then:

1. Reversing the reaction gives: $K' = \frac{1}{K}$

2. Multiplying all stoichiometric coefficients by $n$ gives: $K' = K^n$

3. Adding reactions multiplies equilibrium constants.

So relations like $K_2 = K_1^3$ or $K_1K_2 = 1/3$ can arise only from specific reaction manipulations.

2. Check the listed options

The options as printed are:

• A: $K_1K_2 = \dfrac{1}{3}$
• B: $K_2 = K_1 3$
• C: $K_2 = K_1 - 3$
• D: $K_1K_2 = 3$

There seems to be a formatting issue in option B. It is likely intended to be either: $K_2 = K_1^3$ or something similar.

3. Physical plausibility

A relation such as $K_2 = K_1 - 3$ is not a standard equilibrium-constant transformation law.

Equilibrium constants combine by multiplication, division, and powers, not by subtraction. Therefore option C is almost certainly not correct in a properly stated equilibrium problem.

Among the given choices, the most chemically meaningful type of relation is typically: $K_2 = K_1^3$ if the second reaction is obtained by multiplying the first by $3$.

4. Conclusion

Because the actual reactions are missing, the problem cannot be solved uniquely from the provided statement.

But based on equilibrium rules, option C is not a valid standard relation, so I do not agree with the stored answer.

The most likely intended correct relation is: $K_2 = K_1^3$ if the second reaction is three times the first reaction.