Comparing equilibrium constants between acetone and benzaldehyde in nucleophilic addition reaction
The information given by your teacher is incorrect. The $K_\mathrm{eq}$ value of acetone for cyanohydrin formation (a nucleophilic addition reaction) is less than the $K_\mathrm{eq}$ shown by benzaldehyde
According to Organic Chemistry: Structure, Mechanism, Synthesis [1 , p. 595], the values of $K_\mathrm{eq}$ for cyanohydrin formation are as follows:
$$
\begin{array}{lr} \hline
\mathrm{Compound} &amp; K_\mathrm{eq} \\ \hline
\text{Acetone} &amp; 20 \\
\text{Ethanal} &amp; 10^5 \\
\text{Benzaldehyde} &amp; 210 \\
\textit{p}\text{-Methoxybenzaldehyde} &amp; 30 \\
\text{Acetophenone} &amp; 0.8 \\ \hline
\end{array}
$$
Hence the statement that the $K_\mathrm{eq}$ is greater for acetone is incorrect. This seems accurate considering that acetone is more sterically hindered than benzaldehyde, which is a dominating factor for nucleophilic addition reaction since they are kinetically stable reactions.
This document has a much larger cases of compounds that are compared for multiple reactions, in all of which benzaldehyde is more reactive than acetone.
However, there is one minor issue with your reasoning given in the question.

Benzaldehyde does not have a −M group in it. However if you add another functional group into the ring, then we say that the $\ce{-CHO}$ group acts as a −M group. Actually, $\ce{-CHO}$ has a +R effect from the phenyl ring attached to it which makes the ring in benzaldehyde deactivated. So, the carbon in benzaldehyde is less electrophilic than formaldehyde which has no such effects.
Reference

Ouellette, R. J.; Rawn, J. D. Organic Chemistry: Structure, Mechanism, and Synthesis, 1st ed.; Elsevier, 2014. ISBN: 978-0-12-800780-8 (Google Books Preview)