Common Ion Effect - Ionic Equilibrium
Let's consider an aqueous solution, the concentration of which is $\pu{1 M}$ in $\ce{NH3}$ and $\pu{1 M}$ in $\ce{NaOH}$. Thus, following equilibrium would be taken place:
$$\ce{NH3 (aq) + H2O &lt;=&gt; NH4+ (aq) + OH- (aq)}\tag1$$
$$\ce{NaOH (aq) -&gt; Na+ (aq) + OH- (aq)}\tag2$$
The $\mathrm{p}K_\mathrm{b}$ of equilibrium $(1)$ is 4.75, thus $K_\mathrm{b} = \pu{1.78E{-5}}$. In pure ammonia solution, from equation $(1)$:
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}\tag3$$
If ionized amount at equilibrium is $\alpha$, then
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{\alpha \times \alpha}{1-\alpha} = \alpha^2 \ \Rightarrow \ \therefore \ \alpha = \sqrt{\pu{1.78E{-5}}} = \pu{4.22E{-3}}$$
Assumptions: $\alpha \lt\lt 1$, and thus $1-\alpha \approx 1$, and $\alpha  \gt \gt \pu{1.00E{-7}}$ and autoionization of water can be ignored. At the end, since $[\ce{NH4+}] = [\ce{OH-}] = \alpha = \pu{4.22E{-3}}$, both of these assoumtions are correct.
Now consider, if you have $\ce{NaOH}$ in your solution. Since it is a strong base, it completely dissociate according to equation $(2)$. Thus, there is a common ion in this solution: $[\ce{OH-}] = \pu{1 M}$. Hence, from the equation $(3)$:
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{\beta \times (1+\beta)}{1-\beta} = \beta \ \Rightarrow \ \therefore \ \beta = \pu{1.78E{-5}} $$
Hence ($\alpha \gt \beta$), the ionization amount of $\ce{NH3}$ in presence of the common ion $\ce{OH-}$ is less than that of the solution when no common ions are present.
In similar way, you can prove the ionization amount of $\ce{NH3}$ is larger in the presence of the common ion $\ce{NH4+}$ Using following equilibria:
$$\ce{NH3 (aq) + H2O &lt;=&gt; NH4+ (aq) + OH- (aq)}\tag1$$
$$\ce{NH4Cl (aq) -&gt; NH4+ (aq) + Cl- (aq)}\tag4$$
$$\ce{NH4+ (aq) + H2O &lt;=&gt; H3O+ (aq) + NH3 (aq)}\tag5$$