The heat of atomisation of  $P{H}_{3\left(g\right)}$ is  228K.Cal $mo{l}^{-1}$  and that of $P_2{H}_{4\left(g\right)}$  is 355K.Cal  $mo{l}^{-1}$ The energy of the P-P bond is (in K.Cal);

$\Delta H_1=3\times$  bond energy P-H = 228 K.Cal. $mo{l}^{-1}$
 Bond energy P-H = 76 K.Cal. $mo{l}^{-1}$
Now $\Delta H_2=\left[4\times bondenergyofP-H\right]$$+\left[BondenergyofP-P\right]$
$355=4\left(76\right)+BondenergyofP-P$ 
$\therefore$  Bond energy of P-P = 355-4(76)
 = 51.K.Cal