physical chemistry - Pressure dependency in Haber Bosch ammonia synthesis

Haber-Bosch ammonia synthesis reaction: $$\ce{3H2 +N2 -> 2NH3}$$

According to the ideal gas law: $pV=nRT$ where constant volume implies $\displaystyle \frac nV=\frac p{RT}$

Then $\displaystyle K_\mathrm c=\frac{[\ce{NH3}]^{2}}{[\ce{H2}]^3\cdot [\ce{N2}]}$

When I substitute $\displaystyle \frac nV=c=\frac {p_{\text{gas}}}{RT}$ I get:

$$K_\mathrm c=\frac{\left(\frac{p(NH_3)}{RT}\right)^2}{\left(\frac{p(H_2)}{RT}\right)^3\left(\frac{p(N_2)}{RT}\right)}=\frac{R^2T^2}{p^2}$$

But according to Le Chatelier's principle, increased pressure should result in an increased $K_\mathrm c$, but my $\displaystyle K_\mathrm c\approx \frac 1{p^2}$.

Where is the error? Is is impossible to combine the partial pressures like that because they scale differently with total pressure? Can one derive from the gas law the pressure dependency of this reaction?

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Edit after solved:
First mistake was that the formula had $K_c$ (constant) instead of Q, $$Q=\frac{\left(\frac{p(NH_3)}{RT}\right)^2}{\left(\frac{p(H_2)}{RT}\right)^3\left(\frac{p(N_2)}{RT}\right)}=\frac{R^2T^2}{p^2}$$ where Q is the current value/ratio of concentrations, so the reaction tries to get Q back to $K_c$ so for greater pressure p, Q is smaller so the reaction goes in the direction of NH3...

Answer

How do you know that $p_{\ce{NH_3}}$, $p_{\ce{H_2} }$ and $p_{\ce{N_2}}$ are all same and all equals $p$. You can't just substitute same $p$ for all of them because according to Le-chatterlier's principle, the increase in total pressure of the system, distorts the equilibrium, but your $p$ is not at all total pressure. So, you can't just say anything from your result. By the formula, $$K_p = K_x (P)^{\Delta n}$$ you can say something. Here $P$ is the total pressure of the system. (This is because the formula is derived by using Dalton's law of partial pressure which says, $p_i = x_i\dot P$ where $p_i$ is the partial prssure of the $i^{th}$ gas, $x_i$ is its mole fraction and $P$ is the total pressure of the system.)

Here $\Delta n = -2$, so , $$K_p = K_x/P^2$$ If you now increase $p$ at constant temperature, $K_p$ remains constant as it depends only on temperature. So, $K_x$ also increases to make $K_p$ remain constant, but $$K_x = x^2_{\ce{NH3}}/x^3_{\ce{H2}}\cdot x_{\ce{N2}}=n^2_{\ce{NH3}}\dot N^2/n^3_{\ce{H2}}\cdot n_{\ce{N2}}$$ where, $N = n_\ce{NH3} + n_\ce{N2} + n_\ce{H2}$

$K_x$ increasing means that the moles of $\ce{NH3}$ increase while the moles of $\ce{H2}$ and $\ce{N2}$ decrease, which is the favour of forward reaction and as $$K_p = K_c (RT)^{\Delta n}$$ here $\Delta n = -2$ and the temperature doesn't change so $$K_c= K_p (RT)^2 -> K_c \propto K_p$$ As $K_p$ increases, $K_c$ also increases.