How do you write the the reaction of lead(II) nitrate (aq) with sodium iodide (aq) to form lead (II) iodide precipitate and sodium nitrate solution?

You're dealing with a double replacement reaction in which two soluble ionic compounds in aqueous solution react to form an insoluble solid that precipitates out of solution.

In this case, lead(II) nitrate, #"Pb"("NO"_3)_2#, and sodium iodide, #"NaI"#, both soluble in water, will exist as ions in aqueous solution

#"Pb"("NO"_3)_text(2(aq]) -> "Pb"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)#

#"NaI"_text((aq]) -> "Na"_text((aq])^(+) + "I"_text((aq])^(-)#

When these two solutions are mixed, the lead(II) cations, #"Pb"^(2+)#, and the iodide anions, #"I"^(-)#, will bind to each other and form lead(II) iodide, an insoluble ionic compound.

The other product of the reaction is aqueous sodium nitrate, #"NaNO"_3#, which will exist as ions in solution.

You can thus say that

#"Pb"("NO"_3)_text(2(aq]) + color(red)(2)"NaI"_text((aq]) -> "PbI"_text(2(s]) darr + color(red)(2)"NaNO"_text(3(aq])#

The complete ionic equation, which includes all the ions present in solution, will look like this

#"Pb"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) + color(red)(2)"Na"_text((aq])^(+) + color(red)(2)"I"_text((aq])^(-) -> "PbI"_text(2(s]) darr + color(red)(2)"Na"_text((aq])^(+) + 2"NO"_text(3(aq])^(-)#

The net ionic equation, which eliminates spectator ions, i.e. the ions that are present on both sides of the equation, will look like this

#"Pb"_text((aq])^(2+) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-)))) + color(red)(cancel(color(black)(color(red)(2)"Na"_text((aq])^(+)))) + color(red)(2)"I"_text((aq])^(-) -> "PbI"_text(2(s]) darr + color(red)(cancel(color(black)(color(red)(2)"Na"_text((aq])^(+)))) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-))))#

This will be equivalent to

#"Pb"_text((aq])^(2+) + color(red)(2)"I"_text((aq])^(-) -> "PbI"_text(2(s]) darr#

Lead(II) iodide is a yellow solid that precipitates out of solution.