Is the application of Le Chatelier's principle wrong in this case? In the reaction:
$$\ce{N2(g) + 3H2(g) &lt;=&gt; 2NH3(g)}$$
Suppose the system has 3 moles of $\ce{N2}$, 1 mole of $\ce{H2}$ and 1 mole of $\ce{NH3}$ at equilibrium. $Q_\chi=K_\chi=8.33$ and $Q_p=K_p$.

Now I add 5 moles of $\ce{N2}$ in the container by keeping (i) pressure and (ii) volume constant (temperature constant in both).

According to Le Chatelier's principle:


If pressure is constant: equilibrium shifts to backward direction.
If volume is constant: equilibrium shifts to forward direction.


But if I look at the equilibrium in terms of mole fraction, $Q_\chi$ will be the same in both the cases and hence equilibrium should shift in the same direction i.e. in backward direction.

Why do they predict different equilibrium states?

[$K_\chi$ is equilibrium constant in terms of mole fraction, $K_p$ is equilibrium constant in terms of partial pressures].
Try doing the calculation using the fact that $K_x$ actually is not constant at a given temperature.  It depends on total pressure as well as temperature.  Assuming you maintain ideal gas behavior $K_p$, and only $K_p$, is constant when temperature is fixed.  In terms of that equilibrium constant you should find the reaction following Le Chatelier's Principle both of your cases.