Finding the concentration of each ion in saturated solution of lead(II) hydroxide
The mistake you made is that you assumed mass concentrations of the ions from the stoichiometric relation, which is only applicable to the amount of substance, and, as a consequence, amount concentration.

You have to convert mass concentration $γ$ to the molar concentration c first:

$$c(\ce{Pb(OH)2}) = \frac{γ(\ce{Pb(OH)2})}{M(\ce{Pb(OH)2})} = \frac{\pu{0.155 g L-1}}{\pu{241.21 g mol-1}} = \pu{6.43E-4 mol L-1}$$

And, since it's a saturated solution, then a nearly complete dissociation can be assumed:

$$[\ce{Pb^2+}] = c(\ce{Pb(OH)2}) = \pu{6.43E-4 mol L-1}$$

$$[\ce{OH-}] = 2 × c(\ce{Pb(OH)2}) = 2 × \pu{6.43E-4 mol L-1} = \pu{1.29E-3 mol L-1}$$