It's often helpful to use state symbols when dealing with solutions, especially in this case. Let's say you mix $\ce{KI}$ with $\ce{NaNO_3}$. Because the $\ce{KI}$ and $\ce{NaNO_3}$ are soluble and fully dissolved (or assumed to be) in the solutions, before mixing, we write them as $\ce{KI}$ $\ce{(aq)}$ and $\ce{NaNO_3}$ $\ce{(aq)}$. The aq stands for aqueous. After you pour them together, the ions in both solutions mix into the larger volume. However, they're still fully dissolved, so no reaction happens. Before, you had $\ce{K^+}$ $\ce{(aq)}$ and $\ce{I^-}$ $\ce{(aq)}$ with $\ce{Na^+}$ $\ce{(aq)}$ and $\ce{NO_3^-}$ $\ce{(aq)}$, and that's exactly what you have afterward. ($\ce{KI (aq)}$ means the same as $\ce{K^+}$ $\ce{(aq)}$ + $\ce{I^-}$ $\ce{(aq)}$, and the same with sodium nitrate).
However, with $\ce{KI}$ $\ce{(aq)}$ and $\ce{PbNO_3}$ $\ce{(aq)}$, they are both initially fully dissolved, but after pouring them together, the $\ce{Pb^{2+}}$$\ce{(aq)}$ ions react with the $\ce{I^-}$ $\ce{(aq)}$ ions and form a yellow solid ($\ce{PbI_2}$) that IS NOT SOLUBLE in water. It comes out of solution as a precipitate. Thus, we write it as $\ce{PbI_2}$ (s). Something actually happened: you had ions that mixed to form a solid.
The equation you have is better written this way:
$\ce{KI (aq) + Pb(NO_3)_2 (aq) -> KNO_3 (aq) + PbI_2 (s)}$.
Notice the solid. If a reaction produces only aqueous products, then nothing happens beyond mixing. If it produces anything that's non-aqeuous (like in this case), THEN there's a chemical reaction.