Ok, my data is slightly different than yours but it shouldn't make any significant differences. So I have these bond enthalpies: N-N: 163 kJ mol^{-1}, P-P: 200 kJ mol^{-1}, N triple bond N: 945 kJ mol^{-1} and P triple bond P: 481 kJ mol^{-1}

Now consider P_{4}, there are six P-P single bonds, total bond enthalpy = 1200 kJ mol^{-1}

Compare this to two P_{2} molecules, which have total bond enthalpy of 2 x 481 or using your data 2 x 490 = 980 kJ mol^{-1} which is significantly less than that of P_{4}. Therefore it is more favourable to have P_{4} than two P_{2}

Similar calculation for nitrogen, if there were six N-N bonds the total bond enthalpy would be 978 kJ mol^{-1}. Two N_{2} molecules would have bond enthalpy 945 x 2 = 1890 kJ mol^{-1}. Thus, it is much more stable to have two N_{2} molecules than one N_{4}