It's not too bad...all you do is take what you know and try to manipulate to what you originally want
So, you need NH3 and O2 on the left, NO and H2O on the right
The first given enthalpy has NO on the right, but only 2NO...you need 4NO
So double everything (including the enthalpy)
2N2+2O2 -> 4NO; dH = 361.2 kJ
The second equation has NH3 on the right side, so move it to the left and invert dH
2NH3 -> N2 + 3H2; dH = 91.8 kJ
But you need 4NH3, so again double everything
4NH3 -> 2N2 + 6H2; dH = 183.6 kJ
Lastly, you need 6H2O on the right, the last equation has only 2H2O, so tripple everything
6H2 + 3O2 -> 6H2O; dH = -1316.1 kJ
Add it all together
2N2 + 2O2 -> 4NO; dH = 361.2 kJ
4NH3 -> 2N2 + 6H2; dH = 183.6 kJ
6H2 + 3O2 -> 6H2O; dh = -1316.1 kJ
2N2+2O2+4NH3+6H2+3O2 -> 4NO + 2N2+6H2+6H2O; dH = -771.3 kJ
6H2 and 2N2 are on both sides, so cancel them.
4NH3+5O2 -> 4NO+6H2O; dH = -771.3 kJ
EDIT: Corrected 4NH3+2O2 -> 4NO+6H2O; dH = -771.3 kJ ... other than that the final enthalpy remains the same
dH = -771.3 kJ is the final enthalpy