$\mathop {N{H_4}OH}\limits_{Weak\,base}  \leftrightarrow \mathop {NH_4^ +  + O{H^ - }}\limits_{\Delta H = x\,kJ\,mo{l^{ - 1}}....(ii)} $

${H^ + } + O{H^ - } \to {H_2}O$

             $\Delta H =  - 55.90\,kJ\,mo{l^{ - 1}}....(iii)$

             (from neutralisation of strong acid and strong base)

From equation $(i),(ii)$ and $(iii)$

$N{H_4}OH + HCl \to NH_4^ +  + C{l^ - } + {H_2}O$

         $\Delta H =  - 51.46\,kJ\,mo{l^{ - 1}}$

$\therefore x + ( - 55.90) =  - 51.46$

$x =  - 51.46 + 55.90$

$ =  4.44\,kJ\,mo{l^{ - 1}}$

$\therefore $ Enthalpy of ionisation of 

$N{H_4}OH = 4.44\,kJ\,mo{l^{ - 1}}$