Solving for the momentum from eigenfunctions

When you have a solution to a time-dependent Schrodinger Equation, $$Psi(x,t)=expleft({-frac{ihbar^2k_0^2t}{2m}}right)sin(k_0x), tag{1}$$ and want to know the distribution of momentum
at time t, you can decompose the sin function into $$sin(k_0x)=frac{exp(ik_0x)-exp(-ik_0x)}{2i}, tag{2}$$ and say each is an eigenfunction of $hat{p}$.

I guess the momenta are $frac{exp(ik_0x)}{2i}$ and $frac{-exp(-ik_0x)}{2i}$ with equal probability 1/2 because the coefficients squared ${(1/(2i))}^2$ are equal.

I’m still not sure why the eigenvalue would be $-k_0$, $k_0$ and not $-hbar k_0$, $hbar k_0$.

EDIT: The momentum here means the distribution of momenta. Due some feedback, I decided to attach the original problem, citing Professor Irfan Siddiqi’s exam from 2018:

A free particle of mass m moving in one dimension is known to be in the initial state

$$ psi(x,0)=sin (k_0 x)$$

1. What is $ψ(x,t)$? [10 pts]
2. What value of momentum will a measurement yield at time t, and with what probabilities will these values occur? [10 pts]