WE are given with the solubility constant of lead (II) iodide of 7.1 x10-9 in a 0.5 M kI. Ksp or solubility constant is equal to the product of the concentrations of each ion raised to their respective number of ions. PbI2 when dissociates results to 1 mole Pb2+ and 2 moles I-. The equation is Ksp = (x)(2*0.4)^2 = 7.1 x10-9 where x is the solubility of PbI2. Answer is 1.1 x10^-18 M.
