Anyone knows how to do this?

Initial concentrations :
[H₃PO₄]ₒ = [(98 g) / (98 g/mol)] / (1 L) = 1 M
[NaOH]ₒ = [(59.9955 g) / (39.997 g)] / (1 L) = 1.5 M

Firstly, H₃PO₄ reacts with NaOH to give NaH₂PO₄ and H₂O.
H₃PO₄(aq) + NaOH(aq) → NaH₂PO₄(aq) + H₂O(l)
Obviously, H₃PO₄ is the limiting reactant (limiting reagent) which completely reacts.
After reaction :
[NaOH] = (1.5 - 1) M = 0.5 M
[NaH₂PO₄] = 1 M

Then, NaH₂PO₄ reacts with NaOH to give Na₂HPO₄ and H₂O
NaH₂PO₄(aq) + NaOH(aq) → Na₂HPO₄(aq) + H₂O(l)
Obviously, NaOH is the limiting reactant (limiting reagent) which completely reacts.
After reaction :
[NaH₂PO₄] = (1 - 0.5) M = 0.5 M
[Na₂HPO₄] = 0.5 M

NaH₂PO₄ and Na₂HPO₄ completely dissociates in water.
Then, [H₂PO₄⁻] = [HPO₄²⁻] = 0.5 M

Consider the dissociation of H₂PO₄⁻ ions :
H₂PO₄⁻(aq) + H₂O(l) ⇌ HPO₄²⁻(aq) + H₃O⁺(aq) ...... pKₐ₂ = 7.20

pH = pKₐ₂ - log([H₂PO₄⁻]/[HPO₄²⁻]) = 7.20 - log(0.5/0.5) = 7.20