Thermodynamic stability of compounds can be determined by obviously enthalpy of formation ($\Delta H_{_\mathrm f}$) of individual compounds. The enthalpy of formation will be lesser if the compound is formed from its constituent elements enjoys some greater stability.
Among the above given compounds, considering any extra stability of the compounds, their thermodynamic stability can be guessed.
For phenol, when 1 mole of it is formed there can be slightly weak intermolecular Hydrogen bonding($\ce{C6H5-OH...OH-C6H5)}$. The hydrogen bonding is slightly weak because the lone pair of oxygen atom is delocalised with the benzene ring thus creating a $\delta$+ charge on the $\ce O$ atom, due to which the dipole dipole interaction between $\ce H$ of one $\ce{OH}$ and $\ce O$ of other phenol molecule is weakened. Thus, phenol enjoys only weak intermolecular hydrogen bonding.
For 4-nitrophenol, there is stronger intermolecular hydrogen bonding, which can exist between two phenol molecules as:

Due to enhanced no. of hydrogen bonds, the compound enjoys more stability than ordinary phenol and therefore its enthalpy of formation is more than phenol.
For hydroquinone, the number of hydrogen bonding is even more. One molecule can have a hydrogen bond with four other molecules through four hydrogen bonds as shown:

Thus the molecule of hydroquinone enjoys much more stability due to large number of hydrogen bonding, making its thermodynamic stability even more than p-nitro phenol.
Thus the stability order will be $$\text{hydroquinone > 4-nitrophenol > phenol}$$