Redacted.

bumpp...

Let $g$ be a primitive root $modulo \; p$. Write the sum as: $\sum_{a=1}^{p-1} g^{ak_i}$, and let the $k_i$ assigned to $g^a$ be from now on $x_a$. Thus we need to choose a permutation $x_1, x_2,...x_{p-1}$ of the set $1,2,...p-1$ such that $\sum_{a=1}^{p-1} g^{ax_a}$ is divisible by $p$. Now we see that $ p \; \vert \;g^a + g^b \Rightarrow a \equiv b+\frac{p-1}{2} \mod {p-1}$. And this motivates us letting $x_{n+\frac{p-1}{2}}=x_n+\frac{p-1}{2}$, and looking at the sum $g^{ax_a}+g^{(a+\frac{p-1}{2})x_{a+\frac{p-1}{2}}}$. We want this to be divisible by $p,$ thus we want $ax_a +\frac{p-1}{2} \equiv (a+\frac{p-1}{2})(x_{a}+\frac{p-1}{2}) \mod p-1 $ and after opening the brackets we need $\frac{p-1}{2} \equiv (\frac{p-1}{2})^2 + \frac{p-1}{2}(a+x_a) \mod p-1 $, equivalently: $1 \equiv \frac{p-1}{2}+a+x_a \mod 2$ and this is easy as for $p \equiv 3 \mod4$ picking $x_a=a$ is sufficient and for $p \equiv 1 \mod4$ just $x_a=a+1$ and $x_{p-1}=1$ is enough. $q.e.d$