Answer
The solution will not change visibly, because no precipitation is going to happen.
Work Step by Step
This is the equilibrium reaction:
$Ca(OH)_2(s) \lt -- \gt Ca^{2+} (aq) + 2OH^-(aq)$
When we add $HCl$ to the solution, it will react with the hydroxide ions:
$HCl(aq) + OH^-(aq) -- \gt Cl^-(aq) + H_2O(l)$
Since the $OH^-$ is being consumed, its concentration is going to decrease.
Following the Le Chatelier's principle, if we reduce the concentration of one of the products $(OH^-)$, the equilibrium will try to favor that side, to compensate for that.
Therefore, more $Ca(OH)_2$ would dissolve, producing more ions.
The solution was already saturated (didn't have any precipitate), so that change will have no visual effect.
