Suppose initially you had 1 mole of $\ce{N2}$ and 3 moles of $\ce{H2}$, and that $n$ moles of $\ce{N2}$ had reacted. Then you would have, at equilibrium, $1-n$ moles of $\ce{N2}$, $3(1-n)$ moles of $\ce{H2}$, and $2n$ moles of $\ce{NH_3}$. So the total moles at equilibrium would be $4-2n$, and the mole fractions (equal to the partial pressures in atm) would be $$x_{\ce{N2}}=\frac{1-n}{4-2n}$$$$x_{\ce{H2}}=\frac{3(1-n)}{4-2n}$$ and $$x_{\ce{NH3}}=\frac{2n}{4-2n}$$So the equilibrium relation would be $$\frac{(2n)^2(4-2n)^2}{3(1-n)^4}=K_p$$If we had added $\delta$ moles of inert to the mix, we would have obtained:$$\frac{(2n)^2(4+\delta-2n)^2}{3(1-n)^4}=K_p$$or equivalently $$2n(4+\delta-2n)=\sqrt{3K_p}(1-n)^2\tag{1}$$
ADDENDUM
If we solve Eqn. 1 analytically for $n=n_0$ in the base case $\delta=0$, we obtain:
$$n_0=1-\frac{2}{\sqrt{4+\sqrt{3K_p}}}\tag{2}$$Furthermore, if we differentiate Eqn. 1 with respect to $\delta$ and take the limit as $\delta\rightarrow 0$, we obtain: $$\left(\frac{dn}{d\delta}\right)_{\delta \rightarrow 0}=-\frac{n_0}{2\sqrt{4+\sqrt{3K_p}}}\tag{3}$$Therefore, the conversion decreases as we add inert gas to the reaction mixture.
The overall molar concentration $C_0$ in the cylinder is constant (irrespective) of addition of inert to the mixture), and is given by: $$C_0=\frac{P}{RT}$$where P is the total pressure. Therefore, the concentration of NH3 is equal to the overall molar concentration times its mole fraction: $$C_{NH3}=\frac{2n}{4-2n+\delta}C_0$$If we employ Eqn. 3 to evaluate the derivative of $C_{NH3}$ with respect to $\delta$ at $\delta\rightarrow 0$ we obtain:
$$\left(\frac{dC_{NH3}}{d\delta}\right)_{\delta\rightarrow 0}=\frac{-n_0}{2\left(1+\frac{2}{\sqrt{4+\sqrt{3K_p}}}\right)}C_0$$Therefore, as the number of moles of inert gas increases, not only does the number of moles of ammonia decrease, but so also does the ammonia concentration in the gas mixture.
