The equation that is incorrect is:

First, recall Kohlrausch’s Law of Independent Migration of Ions, which states

$$\Lambda_m^0(\text{electrolyte}) = \lambda^0_+ + \lambda^0_-$$

where $$\lambda^0_+$$ is the limiting molar ionic conductivity of the cation and $$\lambda^0_-$$ is that of the anion. With this law we can expand every limiting molar conductivity appearing in the options into the sum of its two ionic parts and then compare both sides algebraically, term by term.

Option A proposes

$$ (\Lambda_m^0)_{\text{KBr}} - (\Lambda_m^0)_{\text{NaCl}} \;=\; (\Lambda_m^0)_{\text{KBr}} - (\Lambda_m^0)_{\text{KCl}} $$

Using Kohlrausch’s law on each symbol:

$$$ \begin{aligned} (\Lambda_m^0)_{\text{KBr}} &= \lambda_{\text{K}}^0 + \lambda_{\text{Br}}^0,\\ (\Lambda_m^0)_{\text{NaCl}} &= \lambda_{\text{Na}}^0 + \lambda_{\text{Cl}}^0,\\ (\Lambda_m^0)_{\text{KCl}} &= \lambda_{\text{K}}^0 + \lambda_{\text{Cl}}^0. \end{aligned} $$$

Now substitute:

$$$ \begin{aligned} \text{LHS} &= (\lambda_{\text{K}}^0 + \lambda_{\text{Br}}^0) - (\lambda_{\text{Na}}^0 + \lambda_{\text{Cl}}^0)\\[4pt] &= \lambda_{\text{K}}^0 - \lambda_{\text{Na}}^0 + \lambda_{\text{Br}}^0 - \lambda_{\text{Cl}}^0, \end{aligned} $$$

$$$ \begin{aligned} \text{RHS} &= (\lambda_{\text{K}}^0 + \lambda_{\text{Br}}^0) - (\lambda_{\text{K}}^0 + \lambda_{\text{Cl}}^0)\\[4pt] &= \lambda_{\text{Br}}^0 - \lambda_{\text{Cl}}^0. \end{aligned} $$$

The two sides will be equal only if $$\lambda_{\text{K}}^0 - \lambda_{\text{Na}}^0 = 0,$$ i.e. $$\lambda_{\text{K}}^0 = \lambda_{\text{Na}}^0,$$ which is not true experimentally. Therefore option A is incorrect.

Option B states

$$ (\Lambda_m^0)_{\text{KG}} - (\Lambda_m^0)_{\text{KQ}} \;=\; (\Lambda_m^0)_{\text{NaG}} - (\Lambda_m^0)_{\text{NaQ}}. $$

Expanding each term (with G and Q as arbitrary anions):

$$$ \begin{aligned} \text{LHS} &= (\lambda_{\text{K}}^0 + \lambda_{\text{G}}^0) - (\lambda_{\text{K}}^0 + \lambda_{\text{Q}}^0)\\[4pt] &= \lambda_{\text{G}}^0 - \lambda_{\text{Q}}^0,\\[8pt] \text{RHS} &= (\lambda_{\text{Na}}^0 + \lambda_{\text{G}}^0) - (\lambda_{\text{Na}}^0 + \lambda_{\text{Q}}^0)\\[4pt] &= \lambda_{\text{G}}^0 - \lambda_{\text{Q}}^0. \end{aligned} $$$

Both sides are identical, hence option B is correct.

Option C claims

$$ (\Lambda_m^0)_{H_2O} \;=\; (\Lambda_m^0)_{\text{HCl}} + (\Lambda_m^0)_{\text{NaOH}} - (\Lambda_m^0)_{\text{NaCl}}. $$

Writing each conductivity in ionic form:

$$$ \begin{aligned} (\Lambda_m^0)_{H_2O} &= \lambda_{\text{H}}^0 + \lambda_{\text{OH}}^0,\\ (\Lambda_m^0)_{\text{HCl}} &= \lambda_{\text{H}}^0 + \lambda_{\text{Cl}}^0,\\ (\Lambda_m^0)_{\text{NaOH}} &= \lambda_{\text{Na}}^0 + \lambda_{\text{OH}}^0,\\ (\Lambda_m^0)_{\text{NaCl}} &= \lambda_{\text{Na}}^0 + \lambda_{\text{Cl}}^0. \end{aligned} $$$

Substituting into the RHS:

$$$ \begin{aligned} \text{RHS} &= (\lambda_{\text{H}}^0 + \lambda_{\text{Cl}}^0) + (\lambda_{\text{Na}}^0 + \lambda_{\text{OH}}^0) - (\lambda_{\text{Na}}^0 + \lambda_{\text{Cl}}^0)\\[4pt] &= \lambda_{\text{H}}^0 + \lambda_{\text{OH}}^0, \end{aligned} $$$

which matches the LHS exactly. Therefore option C is correct.

Option D proposes

$$ (\Lambda_m^0)_{\text{NaBr}} - (\Lambda_m^0)_{\text{NaI}} \;=\; (\Lambda_m^0)_{\text{KBr}} - (\Lambda_m^0)_{\text{KI}}. $$

Again convert to ionic contributions:

$$$ \begin{aligned} \text{LHS} &= (\lambda_{\text{Na}}^0 + \lambda_{\text{Br}}^0) - (\lambda_{\text{Na}}^0 + \lambda_{\text{I}}^0)\\[4pt] &= \lambda_{\text{Br}}^0 - \lambda_{\text{I}}^0,\\[8pt] \text{RHS} &= (\lambda_{\text{K}}^0 + \lambda_{\text{Br}}^0) - (\lambda_{\text{K}}^0 + \lambda_{\text{I}}^0)\\[4pt] &= \lambda_{\text{Br}}^0 - \lambda_{\text{I}}^0. \end{aligned} $$$

Both sides coincide, so option D is correct.

From these detailed algebraic checks we see that only option A fails the equality test.

Hence, the answer is Option A.