Well, both user140606 and Alex Doe are right. Just gonna add a couple of things here.
As user140606 mentions, you would get a reflected wave with a phase difference of $\pi$ only if the eardrum would be completely rigid (this would mean that the impedance there would be infinite. Hopefully, you covered that in your class. If not, you can have a look at a textbook such as "Fundamentals of Acoustics" by Kinsler et al. In the case where you have some finite impedance (most probably complex) you will get some of the energy reflected with some specific phase relation to the incident wave and some of the energy transmitted with a (possibly different) phase relation.
Now, regarding the specifics of the ear canal and the eardrum, one can see that they form a kind of open-closed tube with the characteristic quarter-wavelength resonance being at around 3KHz-4KHz (this reflects very well on the equal loudness contours, else known as Fletcher-Munson contours). Moreover, it is quite well known nowadays to headphone manufacturers that those resonances are something very prominent. This is why the curve they are targeting for their products is not flat but has the shape of the following image.
For more information on this topic, one could refer to "Listener Preferences for Different Headphone Target Response Curves" by Olive et al. and "Identification and Evaluation of Target Curves for Headphones" by Fleischmann et al.
This, in brief, shows that the eardrum does reflect part of the incident energy because otherwise, you wouldn't get a resonance! According to d'Alembert's solution to the wave equation, you get two travelling waves on opposite directions. The steady-state (modal) solutions are more clear to witness from Bernoulli's solution, but since they are solutions to the same equation, one could conceptually think of them as the same thing (we are somewhat abusing the actual mathematics here, but I hope the conceptual connection is clear), which means that two waves moving in opposite directions will create a standing wave (which is what the resonances are in tubes).
To conclude, you do get some cancellation from the reflected wave, but since the energy reflected is not equal to the incident you won't manage to achieve complete cancellation (or double the amplitude at the antinodes).