The first ionization potential of $\text{Na}$ is $5.1\text{eV}$. The value of electron gain enthalpy of ${\text{Na}}^{+}$ will be:

The first ionization potential of $\text{Na}$ represents the energy required for the following process:
$$\text{Na(g)}\to {\text{Na}}^{+}\text{(g)}+e^{-},\Delta H=+5.1\text{eV}$$
The electron gain enthalpy of ${\text{Na}}^{+}$ represents the energy change for the reverse process:
$${\text{Na}}^{+}\text{(g)}+e^{-}\to \text{Na(g)}$$
Since this is the exact reverse of the ionization process, the energy released will have the same magnitude but an opposite sign:
$$\Delta H=-5.1\text{eV}$$