Solubility equilibrium
Q is known as the ionic product which is defined as product of concentration of ions of the electrolyte, each raised to the power of their coefficients in the balanced chemical equation in solution at any concentration. $\pu{K_{sp}}$ is the same but it is applicable only for a saturated solution.  

To find the solubility for the salt with the given $\pu{K_{sp}}$ value, For $\ce{AgCl}$, we have the equilbrium,
$$ \ce{ AgCl(s) &lt;=&gt; Ag+(aq) + Cl-(aq)}$$
If initally we have $‘a ‘$ moles of AgCl, at equilbrium we will have $’a-s ‘$ and we have $\ce{[Ag+]=[Cl-]=s}$. Hence we get the relation as,
$$ \pu{K_{sp}} = s^2 $$ 
This is the maximum concentration of the ions present in solution at equilbrium.(because $\pu{K_{sp}}$ = Q). 
When Q just crosses $\pu{K_{sp}}$, salt is precipitated out. 

Since they are products of ionic concentration, we can’t directly subract Q and $\pu{K_{sp}}$ to find the amount of salt precipitated.  

I’m just working backwards to get $\ce{K_{sp}}$.

Let mass of $\ce{AgCl}$ precipitated be $’m ‘$ g.
No. of moles precipitated $n = \frac{m}{M}$.  Let the volume be $’V’$ litres,
$\ce{[AgCl]}$ precipitated $x = \frac{n}{V}$  

$$ \ce{Ag+ + Cl- -&gt; AgCl} $$

We know initial concentrations of the ions. Let $\ce{[Ag+]}$ and $\ce{[Cl-]}$ be $’p‘$ and $’q‘$ respectively.
Then solubility product is,
$$ \ce{K_{sp}} = (p-x)(q-x) $$

Hope that makes sense.