!! LONG ANSWER !!
So, you're dealing with a buffer that contains monosodium phosphate, #NaH_2PO_4#, and disodium phosphate, #Na_2HPO_4#.
In aquous solution, the species that are of interest for your buffer are dihydrogen phosphate, #H_2PO_4^(-)#, which will act as a weak acid, and hydrogen phosphate, #HPO_4^(2-)#, which will act as the conjugate base.
Since nothing is actually added to the buffer, you have no net ionic equation to write. Now, because phosphoric acid is a triprotic acid, it will dissociate in three steps according to the following equilibrium reactions
#H_3PO_(4(aq)) + H_2O_((l)) rightleftharpoons H_2PO_(4(aq))^(-) + H_3O_((aq))^(+)#, #" "pK_(a1)#
#H_2PO_(4(aq))^(-) + H_2O_((l)) rightleftharpoons HPO_(4(aq))^(2-) + H_3O_((aq))^(+)#, #" "pK_(a2)#
#HPO_(4(aq))^(2-) + H_2O_((l)) rightleftharpoons PO_(4(aq))^(3-) + H_3O_((aq))^(+)#, #" "pK_(a3)#
Notice that your buffer contains #H_2PO_4^(-)# and #HPO_4^(2-)#, which means that the second equilibrium reaction will be established.
As a result, #K_(a2)# will be the more important acid dissociation constant for your buffer.
Notice that the concentration of #HPO_4^(2-)# is 2 times bigger than the concentration of #H_2PO_4^(-)#; this means that the dominant form of the acid will be #HPO_4^(2-)# and that the solution's pH will be bigger than #pK_(a2)#.
Moreover, because the difference between #K_(a2)# and #K_(a3)# is so significant, the concentration of #PO_4^(3-)# can be neglected.
So, since this is a buffer, you can use the Henderson-Hasselbalch equation to solve for the pH. Since the second equilibrium is set, you'll use #pK_(a2)#, which is equal to
#pK_(a2) = -log(K_(a2)) = -log(6.3 * 10^(-8)) = 7.20#
Therefore,
#pH_"sol" = pK_(a2) + log(([HPO_4^(2-)])/([H_2PO_4^(-)]))#
#pH_"sol" = 7.20 + log((0.29cancel("M"))/(0.13cancel("M"))) = 7.20 + 0.35 = color(green)(7.55)#
