The enthalpy of neutralisation of NH$$_4$$OH with HCl is $$-51.46$$ kJ mol$$^{-1}$$ and the enthalpy of neutralisation of NaOH with HCl is $$-55.90$$ kJ mol$$^{-1}$$. The enthalpy of ionisation of NH$$_4$$OH is

The neutralisation of a strong acid with a strong base is essentially the reaction
$$H^+ + OH^- \rightarrow H_2O$$

The heat evolved in this reaction is called the standard enthalpy of neutralisation of strong acid-strong base and is nearly constant for all such pairs. For the data given:

$$\Delta H_{\text{n,\,NaOH+HCl}} = -55.90\ \text{kJ mol}^{-1}$$

This value therefore represents the enthalpy change for the step
$$H^+ + OH^- \rightarrow H_2O\qquad\bigl(\Delta H = -55.90\ \text{kJ mol}^{-1}\bigr)$$

Now consider the neutralisation of the weak base $$NH_4OH$$ with the strong acid $$HCl$$. Two processes occur:

Case 1:

1. Ionisation of the weak base
$$NH_4OH \rightarrow NH_4^+ + OH^- \qquad(\Delta H_{\text{ion}} = ? )$$

2. Neutralisation of the released hydroxide ion
$$H^+ + OH^- \rightarrow H_2O \qquad(\Delta H = -55.90\ \text{kJ mol}^{-1})$$

The overall measured enthalpy of neutralisation for $$NH_4OH$$ with $$HCl$$ is given as

$$\Delta H_{\text{n,\,NH}_4\text{OH+HCl}} = -51.46\ \text{kJ mol}^{-1}$$

Hence, by Hess’s Law,

$$\Delta H_{\text{n,\,NH}_4\text{OH+HCl}} = \Delta H_{\text{ion}} + (-55.90\ \text{kJ mol}^{-1})$$

Substituting the numerical value,

$$-51.46 = \Delta H_{\text{ion}} - 55.90$$
$$\Rightarrow\ \Delta H_{\text{ion}} = -51.46 + 55.90$$
$$\Rightarrow\ \Delta H_{\text{ion}} = +4.44\ \text{kJ mol}^{-1}$$

The positive sign shows that the ionisation of $$NH_4OH$$ is endothermic.

Therefore, the enthalpy of ionisation of $$NH_4OH$$ is

Option D which is: $$+4.44\ \text{kJ mol}^{-1}$$