Aqueous Chemistry: Ranking Ion Concentration
Let's look at the overall reaction:

$\ce{Na3PO4(aq) + 3AgNO3(aq) -&gt; 3NaNO3(aq) + Ag3PO4(s)}$

As you can see $\ce{Na^+}$ and  $\ce{NO_3^-}$ are spectators ions, they don't participate in the main reaction. Their concentrations, taking into account the dilution, are: 

$\ce{[Na^+]= \frac {3 \times 1.0 \times 100}{200}}= 1.5 \ce{M}$.

${[\ce{NO_3^-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

The net ionic equation is:

$\ce{PO4^{3-}(aq) + 3Ag^+(aq) -&gt;  Ag3PO4(s)}$

The initial concentration of phosphate and silver ions taking into account the dilution are:

${[\ce{PO_4^3-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

${[\ce{Ag^+}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

According to the stoichiometry of the reaction: 3 moles of silver ions react with one mole of phosphate ion. This means that silver ion is a limiting reactant. It will completely react to form the precipitate of silver phosphate. On the other hand, the concentration of phosphate at the end of the reaction will be:
${[\ce{PO_4^3-}]= 0.5-\frac{0.5}{3}=\frac{2\times 0.5}{3}=0.33 \ce{M}}$.

In order to determine exactly the concentration of silver ion, we have to use the solubility product of silver phosphate:
${[\ce{Ag^+}]= \frac{K_{sp}}{[\ce{PO_4^3-}]^3}=\frac{8.89\times10^{-17}}{(0.5/3)^3}}=0.04\times10^{-17}\ce{M}$