How to interpret the results of translation operator acting on a momentum state?

It has no physical effect on a momentum eigenstate, only an overall phase difference:

$$defbra#1{leftlangle#1right|}defket#1{left|#1rightrangle}hat{U}(a)ket{q}=e^{-iqa}ket{q}$$

Note that both $q$ and $a$ are constants. Momentum eigenstates are invariant under translation, up to a phase difference.

Momentum is intimately tied to translation because in position space the momentum operator is a derivative. The derivative can be used to construct ('generate') a translation operator.

For $epsilon$ infinitessimal we have begin{align} f(x+epsilon)&approx f(x)+epsilon f'(x) &=left(1+epsilonfrac{d}{dx}right)f(x) &=(1+iepsilonhat p)f(x) end{align} A finite translation can be constructed by composing $n$ of these infinitessimal translations and taking the limit $nrightarrowinfty$. In this limit keep $nepsilon=1$ begin{align} f(x+a)&=f(x+nepsilon a) &=left(1+frac{ihat p a}{n}right)^nf(x) &equiv e^{ihat pa }f(x) &=(1+ihat pa+frac{(ihat p a)^2}{2!}+dots)f(x) end{align} Note the translation operator has a minus sign before $a$ i.e. $hat U_af(x)=f(x-a)$. The momentum state$|qrangle$, being an eigenstate of momentum, behaves simply under translation. begin{align} hat U_a e^{iqx}&=(1-ihat pa+frac{(-ihat p a)^2}{2!}+dots)e^{iqx} &=(1-iqa+frac{(-iq a)^2}{2!}+dots)e^{iqx} &=e^{-iqa}e^{iqx} end{align} So the translation operator just acts on momentum eigenstates by multiplying them by a phase factor. So this leaves the wavefunction invariant since global phase factors don't change the physical wave function. If you construct a linear combination of eigenstates like in the Fourier transform it will act to shift the entire wavefunction. You can easily check this. $$psi(x)=int dk,tilde psi(k)e^{ikx}hat U_apsi(x)= ?$$

Final note: throughout this answer I used the position representation to make things easier i.e. $f(x)$ instead of $langle x|frangle$. With some effort you could translate this back to Bra-Ket notation. For example $hat U_ae^{iqx}=langle x|hat U_a|qrangle$. But I won't do it here because the general idea still holds.