Calculate initial moles of lead nitrate (don't forget vant Hoff factor), It reacts with nacl to form pbcl2 and nano3 , balance equation, then we have two salts ,as pbcl2 is sparingly soluble so take it's solubility s .
According to final condition,
0.83=1.86( i * moles of nano3+ 3s( there are three ions ) )
We don't take i of pbcl2 as it's a solid .
note that s is molar solubility @Siddhant_Mudholkar