Translation operator and parity operator

(This is taken from Introduction to Quantum Mechanics by D. Griffiths, 3rd edition, Problem 6.18 .)

If a system has inverse symmetry, we know that [$hat{H},hat{Pi}] =0$ where $hat{Pi}$ is the parity operator.

This means that eigenstates of the parity operator are eigenstates of $hat{H}$. Namely:

$f(x) = frac{1}{sqrt{pi hbar} }cos(px/hbar)$

$g(x) = frac{1}{sqrt{pi hbar} }sin(px/hbar)$

This is easily seen by doing $hat{Pi} f(x) =f(x) $ and $hat{Pi} g(x) = -g(x) $.

The problem says that translation operator mixes these two states together, meaning that they must be degenerate.

Question

Show that the translation operator mixes these two states together (f and g), meaning that they must be degenerate.

This is what I did:

Translation operator: $hat{T}u(x) = u(x-a)$
$$hat{T}f(x)= frac{1}{sqrt{pi hbar} }cos(pa/hbar)cos(px/hbar) – frac{1}{sqrt{pi hbar} }sin(pa/hbar)sin(px/hbar) = cos(pa/hbar)f(x)- sin(pa/hbar)g(x). $$

I can see that the states are mixed. But it doesn’t have the same energy as $f(x)$ and $g(x)$. If I do $hat{H}hat{T}f(x) = E_n( cos(pa/hbar)f(x)- sin(pa/hbar)g(x) )$

The eigenvalue $E_n$ is multiplied by a constant. The only way this is true if I say that $E_n( cos(pa/hbar)f(x)- sin(pa/hbar)g(x) ) = E_n w(x) $.

Couldn’t I show that they are degenerate by using the simple fact that $hat{H} f(x) = E_n f(x) $ and $hat{H} g(x) = E_n g(x) $, because [$hat{H},hat{Pi}] =0$ ?