$\operatorname{ord}_p(\sum_{i=0}^{p-1}T(\overline i)^{p-1-k}\psi(\overline i))=?$

Question

Let $Z_p$ denote the p-adic integers. Let $T:\mathbb{F_p}\to Z_p$ be a function with the following properties:

1. $\forall x \in \mathbb{F}_p[\overline {T(x)}=x]$
2. $\forall x \in \mathbb{F}_p[T(x)^p=T(x)]$

Let $\psi: \mathbb{F}_p \to Z_p(\zeta_p)$ have the following properties:

1. $\forall x,y \in \mathbb{F}_p[\psi(x+y)=\psi(x)\psi(y)]$

2. $\exists x \in \mathbb{F}_p[\psi(x) \neq 1]$

Moreover, assume $1 \leq k \leq p-2$.

By the way, the \overline notation indicates reduction mod p.

I want to find the p-adic order of $\sum_{i=0}^{p-1}T(\overline i)^{p-1-k}\psi(\overline i)$

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I have been able to find for the special case $p=3$ and $k=1$ using an ad-hoc method. In the special case, I use the fact that $T(\overline 2)=-1$ and the sum becomes $\psi(1)-\psi(2)=\psi(1)[1-\psi(1)],$ which has p-order $\frac 1 {p-1}=\frac 1 2$ by this theorem. But I don't see how to solve the general case. My professor thinks that the order in the general case should be $\frac k {p-1}$ but, alas, I don't see why.