What is the hybridisation in HCN molecule?

$ce{HCN}$ and $ce{HC#CH}$ are linear, triple bonded, with a $π$ system consisting of two perpendicular $π$ bonds. They would be symmetrical in $ce{HC#CH}$, and slightly distorted in $ce{HCN}$, and they leave two orbitals for the sigma system.

In $ce{HCN}$, we hybridize/combine the two remaining orbitals on the carbon atom to form two bonding orbitals, one to the hydrogen, another to the atom on the other side of the carbon (a $ce{C}$ or an $ce{N}$). The natural first combination is a $50$–$50$ split to form two $mathrm{sp}$ orbitals, one directed to the $ce{H}$ and the other directed toward the $ce{N}$.

That's enough most of the time, but if you get picky, you could point out that the electronegativities of $ce{H}$ and $ce{N}$ are quite different ($ce{H}$ $2.1$, $ce{C}$ $2.5$, $ce{N}$ $3.0$), so the nitrogen will be pulling on its $mathrm{sp}$ bond more than hydrogen pulls on its $ce{sp}$ bond, so the $50$-$50$ split readjusts to maybe $70$-$30$ (did your teacher say $~100%$-$0%$?), meaning that the hydrogen gets less of the carbon orbital (i.e., less $mathrm p$-character, more $mathrm s$-character from the carbon orbitals), while the nitrogen gets more of carbon's $mathrm p$-orbital.

If this is so, the $ce{H}$ atom in $ce{HCN}$ should be more easily removed than a H atom in $ce{HC#CH}$. This appears to be correct: the $pK_mathrm a$ of acetylene is $24$ (which is considered to be quite acidic for a hydrocarbon), whereas the $pK_mathrm a$ of hydrocyanic acid is $9.21$, much more acidic (although a weak acid by any other measure).

Your teacher pointed out something interesting, but the exact split ratio remains to be calculated.