The molar conductivity ($\Lambda_m$) of a strong electrolyte depends on the concentration ($c$) according to Kohlrausch's law, which can be mathematically expressed as $\Lambda_m = \Lambda_m^0 - k\sqrt{c}$, where $\Lambda_m^0$ is the molar conductivity at infinite dilution, and $k$ is a constant. The graph of molar conductivity ($\Lambda_m$) against the square root of concentration ($c^{1/2}$) is a straight line with a negative slope.

The unit of molar conductivity ($\Lambda_m$) is Siemens meter squared per mole ($\text{S cm}^2 \text{mol}^{-1}$). Concentration ($c$) has the unit moles per liter ($\text{mol L}^{-1}$), and thus its square root ($c^{1/2}$) has the unit $\text{mol}^{1/2} \text{L}^{-1/2}$.

The slope of the plot of $\Lambda_m$ against $c^{1/2}$ is derived from the expression $k$ in $k\sqrt{c}$. Therefore, to find the correct unit of the slope, we look at the division of the unit of $\Lambda_m$ by the unit of $c^{1/2}$:

$\frac{\text{S cm}^2 \text{mol}^{-1}}{\text{mol}^{1/2} \text{L}^{-1/2}} = \text{S cm}^2 \text{mol}^{-1-1/2} \text{L}^{1/2} = \text{S cm}^2 \text{mol}^{-3/2} \text{L}^{1/2}$

Thus, the correct unit for the slope of a plot of molar conductivity against the square root of concentration for a strong electrolyte is $\text{S cm}^2 \text{mol}^{-3/2} \text{L}^{1/2}$, which matches with Option D.