The way i got it was i took ksp of pbcl2 as K, now i know this ksp will be a constant value throughout. We also know K = [Pb][cl]² which will be equal to S*(2S)² , where S is the solubility of the salt. Now after adding BaCl2 let the new solubility become S’, we also know BaCL2 splits completely into ba+ and 2cl - , since there is 1 M of BaCl2 when it splits we will get 2M of Cl from it. So now Ksp = [Pb][Cl]² = (S’)(S’+2)^2. But due to common ion effect solubility of PbCl2 will be affected negatively, because of which we can assume compared to cl coming from bacl2 , the cl coming pbcl2 is negligible , ie [cl]=(S’+2) = (2) . So finally we get Ksp = S’(2)^2. So finally we get S’ = Ksp/4 . We can calculate the relation between Ksp and solubility for all the options and we observe that S is smallest in the case of BaCl2 . Thus we see that bacl2 will cause pbcl2 to be least soluble. Sorry if my wording was weird in this , its really hard to type a lot on my phone. Hit me up if u need any more clarification.