The connected finite kernel $H$ is not solvable, provided that $n>2$ or $p>2$, see the edit below.

$\def\eps{\varepsilon} \def\m{\mathfrak{m}}$Suppose by contradiction that $H$ is solvable and $m$-th derived group of $H$ is trivial (where $[H,H]$ is the *first* derived group). It implies that for any $k$-algebra $R$ the $m$-th derived group of the abstract group $H(R)$ is also trivial(because for any $G$ we have $[G(R), G(R)]\subset D(G)(R)$). Put $l=2^{m+1}$ and consider $R=k[\eps_1,\dots,\eps_l]/(\eps_1,\dots,\eps_l)^{(p)}$ where $(p)$ means taking $p$-th powers of all the elements of the ideal. Put also $\m=(\eps_1,\dots,\eps_l)$ to be the maximal ideal. I claim that $D(D(...D(H(R)))\neq 1$ ($D$ iterated $m$ times). Indeed, $H(R)=1+\m\mathrm{Mat}_n(R)$. Now note that for any two matrices $A, B\in \mathrm{Mat}_n(R)$ we have (sums represented by dots are actually finite) $$[1+\eps_i A,1+\eps_j B]=(1+\eps_i A)(1+\eps_j B)(1-\eps_i A+\eps^{2}_iA^2-\dots)(1-\eps_j B+\eps_j^2 B^2-\dots)=1+\eps_i\eps_j(AB-BA) \,\,(\rm{mod}\,\, \m^3)$$
Now pick a tuple $A_1,\dots A_{2^{m+1}}$ of elements of the Lie algebra $\mathfrak{gl}_n$ over $k$ such that $(m+1)$ times nested commutator of them is $B\neq 0$. The $(m+1)$ times nested commutator of the elements $1+\eps_iA_i\in H(R)$ then equals $1+\eps_1\dots\eps_{2^{m+1}}B$ modulo $\m^{2^{m+1}+1}$ and thus is not equal to $1$.

Edit: This argument uses that the Lie algebra $\mathfrak{gl}_n$ over a field of characteristic $p$ is not solvable. This is true if $n>2$ or $p>2$, but $\mathfrak{gl}_2$ in characteristic $2$ is actually solvable. This is because $\mathfrak{sl}_2=\ker(\mathfrak{gl_2}\xrightarrow{tr}k)$ has non-trivial center, spanned by the diagonal matrix $\mathrm{diag}(1,1)$, and its quotient by the center is abelian.