What exactly causes equilibrium to shift when pressure is decreased?
This happens because the equilibrium constant $K_p$ is independent of the pressure, although it does depend on temperature, which we will assume to be constant. The equilibrium constant can be written as 

$$\displaystyle K_p =\frac{p_A^2}{p_Np_H^3}$$

where for simplicity I have abbreviated the chemical names. The partial pressure $p$ can be written in terms of the degree of dissociation and total pressure $P_T$, i.e. $p=aP_T$ where $a$ is a mole fraction term in the degree of dissociation. As partial pressure is proportional to the degree of dissociation and total pressure, for example, $p_A =a_A^2  P_T^2$ then the equilibrium constant 

$$K_p=\frac{a_A^2}{a_Na_H^3} \frac{1}{P_T^2}$$

where the terms in $a$ we have not yet evaluated and contain the degree of dissociation $\alpha$. However, you can see that as $K_p$ is constant if $P_T$ decreases then $1/P_T^2$ increases and so  $\displaystyle \frac{a_A^2}{a_Na_H^3} $ must decrease. This means that the degree of dissociation decreases and so the equilibrium 'moves to the left'.

Working out the partial pressure for this reaction is a bit messy. 

Let $\alpha $ be the degree of dissociation then $(1-\alpha)$ of nitrogen reacts , $3(1-\alpha)$ hydrogen and $2\alpha$ ammonia is produced. The total of these is $6-2\alpha$ and so the mole fraction of ammonia is, for example, $a_A=\displaystyle \frac{2\alpha}{6-2\alpha}$ and the partial pressure $p_A=\displaystyle \frac{2\alpha}{6-2\alpha}P_T$. You can make a similar expression for the other species and substitute to get the mole fraction term $\displaystyle \frac{a_A^2}{a_Na_H^3}=\frac{16\alpha^2(\alpha-3)^2}{27(\alpha-1)^4}$. This expression increases as $\alpha$ increases. As you see the expression is messy but the main point is that as the pressure decreases the mole fraction term must decrease which it does by reducing $\alpha$.