Confusion on ion product of water
It is not a chemical problem, but a trivial mathematical problem.

If you have an equation $x \cdot y = c$, where $x$, $y$ are variables and $c$ is constant, then if $x$ increases, $y$ must decrease, otherwise $c$ is not a constant.

The key part is to understand that  chemical equilibrium means existence of 2 ongoing opposite chemical reactions of the same rate. Like

$$\ce{H2O &lt;=&gt; H+ + OH-}$$

where the rate of ion creation equals the rate of ion recombination.

If there is an excess or deficit  of either of  $\ce{H+}$   or $\ce{OH-}$ ions, the rate of their recombination changes, while the rate of their creation remains the same. As consequence, the product of their concentrations converges quickly towards $K_\mathrm{w}$ to be in the equilibrium again.

The ammonia reacts :

$$\ce{NH3 + H+ &lt;=&gt;&gt; NH4+}$$

That creates deficit of $\ce{H+}$. As consequence, dissociation

$$\ce{H2O -&gt; H+ + OH-}$$

Is faster then recombination

$$\ce{ H+ + OH- -&gt; H2O}$$

An alternative reaction mechanism is 
ammonia reacting with water:

$$\ce{NH3 + H2O &lt;=&gt; NH4+ + OH-}$$

what directly produces the excess of $\ce{OH-}$ that recombines with the most of the present $\ce{H+}$.

The production of $\ce{OH-}$ and elimination of $\ce{H+}$ continues, until their concentrations satisfy both the basicity constant of $\ce{NH3}$ ( or equivalently acidity constant of $\ce{NH4+}$ ) and the ion product of water  $K_\mathrm{w}$ 

$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$ 

$$K_\mathrm{b, \ce{NH3}}=\frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}=\frac{[\ce{NH4+}]K_\mathrm{w}}{[\ce{NH3}] [\ce{H+}] }=\frac{K_\mathrm{w}}{K_\mathrm{a, \ce{NH4+}}}$$