Let us fix an odd prime $\require{cancel}p$, and let $\bar{a}_n$ be the reduction of $a_n$ modulo $p$, i.e., as members of $\mathbb{Z}/p=\mathbb{F}_p$. There is a standard trick of letting
$$
\mathbf{b}_n=\begin{bmatrix}\bar{a}_{n+1}\\\bar{a}_n\end{bmatrix}\in\mathbb{F}_p^2,
$$
which converts the second order difference equation into a first-order difference equation (at the cost of having a vector equation rather than a scalar one):
$$
\mathbf{b}_{n+1}=T\,\mathbf{b}_n,\quad T:=\begin{bmatrix}k&1\\1 & 0\end{bmatrix},\quad
\mathbf{b}_0=\begin{bmatrix}1\\0\end{bmatrix}.
$$
So $\mathbf{b}_{n+1}$ depends only on $\mathbf{b}_n$. Note that $\det T=-1$ so $T^{-1}$ exists.
By pigeonhole, there exists $0\leq i<j\leq p^2$ such that $\mathbf{b}_i=\mathbf{b}_j$. Applying $T^{-i}$, you get $\mathbf{b}_{j-i}=\mathbf{b}_0$, and applying $T^n$ gives $\mathbf{b}_{n+j-i}=\mathbf{b}_n$ for all $n$.
So $T(p)$ exists and is equal to the minimal period of $(\mathbf{b}_n)_{n\geq 0}$. (This implies existence of $m(p)$, which is clearly $\leq T(p)$ by putting $j=0$ in the definition.)
For (i):
we consider what $\mathbf{b}_{m(p)},\mathbf{b}_{2m(p)},\mathbf{b}_{3m(p)},\dots$ could be.
The last coordinate must be $0$, but the first coordinate isn't fixed.
However, we know
Claim: $$a_{km(p)+1}\not\equiv 0\pmod{p}$$
Proof.
If $a_{km(p)+1}\equiv 0\pmod{p}$, then $\mathbf{b}_{km(p)}=\mathbf{0}$.
Applying $T^{-km(p)}$ gives $\mathbf{b}_0=\mathbf{0}$, contradiction. $\square$
We have:
$T^{m(p)}\mathbf{b}_0=\bar{a}_{m(p)+1}\mathbf{b}_0$, so
$$T^{m(p)(p-1)}\mathbf{b}_0=(\bar{a}_{m(p)+1})^{p-1}\mathbf{b}_0=\mathbf{b}_0$$
by Fermat's little theorem, and hence $T(p)\leq m(p)(p-1)$.
For (ii):
If $T(p)=m(p)(p-1)$, then $(\bar{a}_{m(p)+1})^j$ for $j=1,2,\dots,p-1$ are all distinct, i.e., $\mu=\bar{a}_{m(p)+1}$ is a generator of $\mathbb{F}_p^\times$. We have $\{\mu,\mu^2,\dots,\mu^{p-1}\}=\{1,2,\dots,p-1\}$ and so applying Wilson's theorem gives $\prod_{k=1}^{p-1}\mu^k=(p-1)!=-1$.
Hence
\begin{align}\prod_{\substack{1\leq j\leq T(p)-1\\m(p)\nmid j}}\bar{a}_j &=\prod_{k=1}^{p-1}\prod_{j=1}^{m(p)-1}\bar{a}_{km(p)+j}\\ &=\prod_{k=1}^{p-1}\prod_{j=1}^{m(p)-1}(\mu^k\bar{a}_j)\\ &=\left(\prod_{k=1}^{p-1}\prod_{j=1}^{m(p)-1}\mu^k\right)\left(\prod_{k=1}^{p-1}\prod_{j=1}^{m(p)-1}\bar{a}_j\right)\\ &=\left[\prod_{j=1}^{m(p)-1}\left(\prod_{k=1}^{p-1}\mu^k\right)\right]\cancelto{1}{\color{red}{\left(\prod_{j=1}^{m(p)-1}\bar{a}_j\right)^{p-1}}}\\ &=(-1)^{m(p)-1} \end{align}
as claimed.
