Both answers are correct.
Let:
$$
f_1(A,B,C) = AB+AC+\overline{C}\\
f_2(A,B,C) = A+\overline{C}
$$
Let's build the thruth table:
$$\begin{array}{|c|c|c|c|c|}
\hline
A & B & C & f_1 & f_2 \\
\hline
0 & 0 & 0 & 1 & 1\\
\hline
0 & 0 & 1 & 0 & 0\\
\hline
0 & 1 & 0 & 1 & 1\\
\hline
0 & 1 & 1 & 0 & 0\\
\hline
1 & 0 & 0 & 1 & 1\\
\hline
1 & 0 & 1 & 1 & 1\\
\hline
1 & 1 & 0 & 1 & 1\\
\hline
1 & 1 & 1 & 1 & 1\\
\hline
\end{array}$$
As you can see the two functions correspond.
Please note that this tabular method of proving that two functions are the same is perfectly valid and is called Proof by exhaustion.
It appears there definitely is some inconsistency on the definition of a dynamic hazard.
Although I have seen some definitions of dynamic hazards that are similar to yours, for example this one:
"Dynamic hazards occur when the output signal has the potential to change more than once when it is expected to make a single transition from 0 to 1 or 1 to 0."
I have also seen the definition of a dynamic hazard simply defined by multiple transitions, and no requirement that the final state be different (it may or may not be).
For example, in this PowerPoint presentation:
"Dynamic hazards: The output could change more than once during input transitions - caused by multiple paths with different delays from input to the output."
So no mention of whether the output has to finally change state or not.
(whereas static hazards are defined in the same paper to be caused by two input combinations that differ in only one variable and produce a single glitch)
Here is another definition of a dynamic hazard from a University of Surrey logic course:
"A dynamic hazard is the possibility of an output changing more than once as a result of a single input change"
Again, no mention of whether the output has to finally change state or not.
So for these definitions, the difference between static and dynamic hazards is defined by the number of unwanted glitches; static 0 and static 1 hazards have only one, and dynamic hazards have more than one.
Using either of the latter two definitions, then in your case of multiple transitions while the output is expected to remain unchanged, would be classed as a dynamic hazard.
Best Answer
The answer is that there is not enough information in the question to decide between the two possibilities.
If you assume that \$\overline{A}\$ is generated by an inverter fed from \$A\$, that the gates have the same propogation delay as each other and the same propagation delay under all conditions then your teacher is right.
When \$A\$ goes from 0 to 1, \$A\$ and \$\overline{A}\$ will both briefly be 1 at the same time, the output of the first level gates will therefore always contain at least one zero and there will be no glitch in the final output.
On the other hand when A goes from 1 to 0, \$A\$ and \$\overline{A}\$ will both briefly be 0 at the same time causing all three of the first stage gates to briefly output a 1 at the same time causing a glitch in the final output.
But those are IMO unwarranted assumptions. Propagation delays vary between gates, between different inputs on the same gate and with the direction of the transition. Depending on how other logic in the system is designed \$A\$ may in fact be generated from \$\overline{A}\$ rather than vice versa.
In the more general case where signals arriving on different paths could arrive in any order there could be a glitch in either direction (or if you get really unlucky a glitch in both directions).