The pH of the phosphate buffer will be #"6.39"#.
The reactions that concern you are
#NaH_2PO_(4(aq)) rightleftharpoons Na_((aq))^(+) + H_2PO_(4(aq))^(-)#, and
#Na_2HPO_(4(aq)) rightleftharpoons 2Na_((aq))^(+) + HPO_(4(aq))^(2-)#
So, adding monosodium phosphate will increase the concentration of #H_2PO_4^(-)#, while adding dosodium phosphate will increase the concentration of #HPO_4^(2-)#.
The first thing you need to do is figure out the concentrations of the two species in the new solution. Starting with monosodium phosphate
#n_("mono") = C * V = "80.0 mmol/L" * 250 * 10^(-3)"L" = "20.0 mmol"#
The volume of the new solution will be
#V_("sol") = "250 mL" + "250 mL" = "500 mL"#, which means that the concentration of monosodium phosphate (and of #H_2PO_4^(-)#) will be
#C_("mono") = n_("mono")/V_("sol") = ("20.0 mmol")/(500*10^(-3)"L") = "40.0 mmol/L"#
Now for the disodium phosphate
#n_("di") = C * V = "30.0 mmol/L" * 250*10^(-3)"L" = "7.50 mmol"#
The concentration of disodium phosphate (and of #HPO_4^(2-)#) in solution will be
#C_("di") = n_("di")/V_("sol") = ("7.50 mmol")/(500*10^(-3)"L") = "15.0 mmol/L"#
Now use the Hendesron-Hasselbalch equation to determine the pH
#pH_("sol") = pKa + log(([HPO_4^(2-)])/([H_2PO_4^(-)]))#
#pH_("sol") = 6.82 + log(("15.0 mmol/L")/("40.0 mmol/L"))#
#pH_("sol") = 6.82 - 0.426 = 6.39#
