Solubility products figure prominently in equilibria involving precipitated species. Following is an example that may arise in a practical situation.
Problem: We have 0.01 M ferrous ion in water and we propose adding a base to drive out the iron as hydroxide. We do not want excess base dissolving into the water so we will try magnesium hydroxide. How well will it work?
The proposed reaction is then
$\ce{Fe^{2+} + Mg(OH)2(s) <=> Mg^{2+} + Fe(OH)2(s)}$
$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}$
Compare this equilibrium constant with:
$K_{sp}(\ce{Mg(OH)2})=[\ce{Mg^{2+}}][\ce{OH^-}]^2=5.61×10^{-12}$ (source)
$K_{sp}(\ce{Fe(OH)2})=[\ce{Fe^{2+}}][\ce{OH^-}]^2=4.87×10^{-17}$ (Ibid.)
So then
$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}=\frac{K_{sp}(\ce{Mg(OH)2})}{K_{sp}(\ce{Fe(OH)2})}=1.15×10^5$
Since the reaction stoichionmetry implies that the sum $[\ce{Mg^{2+}}]+[\ce{Fe^{2+}}]$ will be constant at the original iron concentration of 0.01 M, we find that the iron level remaining in the water may be reduced by five orders of magnitude, the equilibrium concentration is below $10^{-7}$ molar!
Magnesium hydroxide, despite being supposedly "weak" because of its limited solubility, is actually like a strong base here, displacing the weak iron hydroxide base almost quantitatively -- but, in effect, self-regulating because of that limited intrinsic solubility. The same property applies with respect to most heavy metals commonly found in water; their hydroxide solubilities, like that of iron, are much less than magnesium's. Thus magnesium hydroxide (or, in practice, magnesium oxide which becomes the hydroxide in situ) is a good agent for treating water to remove heavy metals. See here, under "Applications".
