80 mol of phosphorous acid, H3PO3, is strongly heated until it has disproportionated completely into a hydride gas and a non-volatile oxoacid. The…

On heating, phosphorous acid disproportionates: $4\,\mathrm{H_3PO_3}\rightarrow3\,\mathrm{H_3PO_4}+\mathrm{PH_3}\uparrow$. Phosphorus goes from $+3$ to $+5$ in phosphoric acid and to $-3$ in phosphine. $\mathrm{PH_3}$ escapes, so only phosphoric acid remains. From $0.80$ mol of $\mathrm{H_3PO_3}$ the amount of $\mathrm{H_3PO_4}$ formed is $0.80\times\tfrac34=0.60$ mol. Phosphoric acid is tribasic (three ionisable $\mathrm{P{-}OH}$ groups), so it needs $3\times0.60=1.8$ mol NaOH, which is $1.8\ \mathrm{L}$ of $1.0\ \mathrm{M}$ solution. Treating the starting acid as dibasic ($0.8\times2=1.6$ mol) ignores the disproportionation, while $2.4$ would count three protons from every mole of starting acid. $0.6\ \mathrm{L}$ treats $\mathrm{H_3PO_4}$ as monobasic.