As a complement to another answer, I'll demonstrate moving from the coordinate representation (wave function) to the momentum representation.
Recalling that
$$\Psi(x,t) = \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$
and putting this expression into the (coordinate representation of the) TDSE, we have
$$i\hbar\frac{\partial}{\partial t}\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) + V(x)\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$
We can move the partials inside the integrals but we must be careful with the potential. Noting that
$$\Phi(p,t) = \int dx \frac{e^{-i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \Psi(x,t)$$
we have
$$V(x)\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) = \int dp\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \int dx'\frac{e^{-i\frac{p}{\hbar}x'}}{\sqrt{2\pi\hbar}}V(x')\Psi(x',t)$$
But,
$$ \int dx'\frac{e^{-i\frac{p}{\hbar}x'}}{\sqrt{2\pi\hbar}}V(x')\Psi(x',t)= V(p) * \Phi(p,t)$$
where $*$ denotes convolution. Thus, we can write
\begin{align}
\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\left(i\hbar\frac{\partial}{\partial t}\Phi(p,t)\right) &= -\int dp \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\left(\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)\right)\\&\qquad + \int dp\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\left( V(p) * \Phi(p,t)\right)
\end{align}
leading to
$$\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \left\{ i\hbar\frac{\partial}{\partial t}\Phi(p,t) = -\frac{\hbar^2}{2m}\left(-\frac{p^2}{\hbar^2}\Phi(p,t) \right) + V(p)*\Phi(p,t)\right\}$$
and so
$$i\hbar\frac{\partial}{\partial t}\Phi(p,t) = \frac{p^2}{2m}\Phi(p,t) + V(p)*\Phi(p,t)$$
Question: do degenerate states always have the same energy?
The answer depends on what one means by degenerate. This word typically means two or more states with the same energy, in which case the answer is trivially yes.
On the other hand, it seems that according to your professor, degenerate means two or more states that share the eigenvalue with respect to some operator. This definition is not particularly useful, because all states are eigenvectors of the identity, with eigenvalue one, which means that all states are degenerate. And even if we restrict ourselves to operators other than the identity, the notion of degenerate is still trivial, because one may consider arbitrary projectors of the form $$P=|\varphi_1\rangle\langle\varphi_1|+|\varphi_2\rangle\langle\varphi_2|$$
which would make $\varphi_1,\varphi_2$ degenerate, for any pair of states $\varphi_1,\varphi_2$. In this sense, this definition of degenerate is vacuous.
Finally, if by degenerate we mean eigenvectors that share the eigenvalue with respect to a particular operator $A$ (fixed, i.e., not arbitrary), then the question becomes more interesting: given a pair of states $a_1,a_2$, such that
$$
\begin{aligned}
A|a_1\rangle&=a|a_1\rangle\\
A|a_2\rangle&=a|a_2\rangle
\end{aligned}
$$
then do we necessarily have
$$
\begin{aligned}
H|a_1\rangle&=E|a_1\rangle\\
H|a_2\rangle&=E|a_2\rangle
\end{aligned}
$$
for some energy $E$?
The answer is not necessarily, as can be seen by constructing simple counter-examples.
Best Answer
The question as I understand it is asking "Why is the eigenvalue of the same form for all of the 1D eigenfunctions I can think of?".
The simple answer is that you can change the basis used to represent the eigenfunctions but the eigenvalue remains invariant under such transformations.
For example, consider
\begin{align} \hat{p} \psi = p\psi \end{align}
where $\psi=\frac{1}{\sqrt{2\pi\hbar}}\exp{ipx/\hbar}$ are the normalised momentum eigenfunctions in position representation, then apply some basis transformation so that $\psi'= \hat{U}\psi$ and $\hat{p}' = \hat{U}\hat{p}\hat{U}^{-1}$ then the eigenfunction equation becomes
\begin{align} \hat{U}\hat{p}\hat{U}^{-1}\psi' = p\hat{U}\hat{U}^{-1}\psi' = p\psi' \end{align}
Therefore, the eigenvalue remains the same under a change of basis i.e. there are multiple eigenfunctions with the same eigenvalue, as you describe in your question.