organic chemistry - Acidity order of nitrophenols

As in other posts I have seen and learnt that, if one is given four compounds, namely para-nitrophenol, ortho-nitrophenol, meta-nitrophenol and phenol and is told to arrange them in order of acidity, then one has to take into account the $\text{-R}$ and $\text{-I}$ effect.

Now, for the inductive effect, the greatest intensity is observed at the ortho-position, followed by meta-, and then para-.

For the resonating effect, the intensity is greatest for para and lowest for ortho, while meta has no such effect. Now keeping in mind, that $\text{-R}$ and $\text{-I}$ both tend to withdraw electrons away from the phenol group (and thus increases the acidity), and the $\text{R}$ effect has more contribution than the $\text{I}$ effect, we can thus deduce that para-nitrophenol is most acidic, followed by ortho-nitrophenol, then meta-nitrophenol and since phenol has no ring activating group, it comes out to be least acidic in the given compounds.

The facts are the proof of my theory, where the $\mathrm{p}K_\mathrm{a}$ for: $$\begin{array}{c|c} \text{Compound} & \mathrm{p}K_\mathrm{a} \\ \hline para\text{-nitrophenol} & 7.16\\ ortho\text{-nitrophenol} &7.2\\ meta\text{-nitrophenol} & 9.3\\ \text{phenol} & 9.95 \end{array}$$

So far so good.

But one of my colleagues, when I asked the problem, tried to give the theory in different way. He said, in para-nitrophenol, hydrogen bonding is strongest, so it can't be most acidic, which in his view port should be least. I knew this is not the case, but I just can't explain or point out where he's been wrong. Can anyone of you make it little easy for me to understand it?