How do I write the complete and net ionic equations for Insoluble compounds?
While lead(II) iodate is indeed sparingly soluble in water, this is not true for sodium hydroxide (111 g/100 mL at 20 °C). The reaction that occurs is a precipitation:

$$\ce{Pb(IO3)2_{(aq)} + 2NaOH_{(aq)} -&gt; Pb(OH)2\downarrow + 2NaIO3_{(aq)}}$$

Both lead salts are not readily soluble in water; however, $\ce{Pb(OH)2}$ has a lower solubility ($K_{sp}=1.43 \times 10^{-20}$) than $\ce{Pb(IO3)2}$ with $K_{sp}=3.69 \times 10^{-13}$. Adding $\ce{NaOH}$ to an aqueous solution of lead iodate should therefore precipitate lead hydroxide, which dissolves again with excess $\ce{NaOH}$ under formation of hydroxoplumbates.