Answer
The equilibrium would favor the formation of $PbCl_2$, due to the common ion effect.
Work Step by Step
This is the equilibrium reaction:
$PbCl_2(s) \lt -- \gt Pb^{2+} (aq) + 2Cl^-(aq)$
When we add $NaCl$, the common ion effect (because it has $Cl^{-}$ ions) will act, and the reaction will be more reactant-favored.
That follows the Le Chatelier's principle. Since we are raising the concentration of one of the products, the equilibrium will try to reduce this concentration, to reduce the change.
