What is the number of representative particles in 4.80 * 10^-3 mol NaI?

$9.6 \times {10}^{-} 3$ total moles of ions.
A sodium iodide unit is conceived to derive from one $N {a}^{+}$ and one ${I}^{-}$. Because matter is electrically neutral these combine in a 1:1 fashion.
You specified $4.80 \times {10}^{-} 3$ $m o l$ of $N a I$. In this molar quantity there are $4.80 \times {10}^{-} 3$ $m o l$ of $N {a}^{+}$ ions, and $4.80 \times {10}^{-} 3$ $m o l$ of ${I}^{-}$; i.e. $9.6 \times {10}^{-} 3$ of individual ions.