n-factor of $H_{3}P{O}_2$ during its diproportionation is $3H_{3}P{O}_2\to P{H}_3+2H_{3}P{O}_3$
 

$$n-factor$$ measures the change in electrons taking place per molecule. In the given reaction, oxidation state of P in each compound is:

$3{\mathrm{H}}_3\stackrel{+1}{\mathrm{P}}{\mathrm{O}}_2\to \stackrel{-3}{\mathrm{P}}{\mathrm{H}}_3+2{\mathrm{H}}_3\stackrel{+3}{\mathrm{P}}{\mathrm{O}}_3$

Half cell reaction:

$$\begin{gathered} {\mathrm{H}}_3{\mathrm{PO}}_2\to {\mathrm{H}}_3{\mathrm{PO}}_3+2{\mathrm{e}}^{-} \left(\mathrm{oxidation}\right) \\ 2{\mathrm{H}}_3{\mathrm{PO}}_2\to 2{\mathrm{H}}_3{\mathrm{PO}}_3+4{\mathrm{e}}^{-} \left(\mathrm{reduction}\right) \\ 3{\mathrm{H}}_3{\mathrm{PO}}_2+4{\mathrm{e}}^{-}\to {\mathrm{PH}}_3+2{\mathrm{H}}_3{\mathrm{PO}}_3+4{\mathrm{e}}^{-} (\mathrm{overall} \mathrm{reaction}) \end{gathered}$$

$$n-factor$$ is calculated as:

$\mathrm{n}-\mathrm{factor}=\frac{\mathrm{number} \mathrm{of} \mathrm{electrons} \mathrm{involved}}{\mathrm{number} \mathrm{of} \mathrm{molecules} \mathrm{involved}}=\frac{4}{3}=1.33$