= Solution
Because $K$ has <characteristic> $p$, the <prime field> $\mathbb F_p$ lies in $K$, and
$$
f(\alpha+c)=(\alpha+c)^p-(\alpha+c)+a=f(\alpha)+c^p-c=0
$$
for every $c\in\mathbb F_p$. These are all $p$ roots. Moreover $f'(t)=-1$, so $f$ is a <separable polynomial>.
Let $m(t)$ be the <minimal polynomial of an algebraic element> of $\alpha$ over $K$. Translation by $c\in\mathbb F_p$ sends an irreducible factor $g(t)$ of $f(t)$ to the irreducible factor $g(t-c)$. The <additive group> of $\mathbb F_p$ therefore acts on the irreducible factors, and every orbit has size $1$ or $p$. An orbit of size $p$ would already contribute $p\deg m$ to the degree-$p$ polynomial $f$, forcing $\deg m=1$. Then $\alpha\in K$, all roots $\alpha+c$ lie in $K$, and $L=K$, contrary to the hypothesis.
Thus $m(t-c)=m(t)$ for every $c\in\mathbb F_p$. If $1\leq d=\deg m<p$, comparison of the coefficient of $t^{d-1}$ in $m(t-c)-m(t)$ for $c\ne0$ gives $-dc$ times the nonzero leading coefficient, a contradiction. Hence $\deg m=p$, and since $m$ is a monic divisor of the monic degree-$p$ polynomial $f$, we have $m=f$. Thus \b[$f$ is irreducible over $K$].
All roots $\alpha+c$ already belong to $K(\alpha)$, so the <splitting field> is
$$
\boxed{L=K(\alpha)}.
$$
As the splitting field of a separable polynomial, $L/K$ is a <Galois extension>. For each $c\in\mathbb F_p$, irreducibility gives a distinct $K$-automorphism
$$
\sigma_c(\alpha)=\alpha+c,
\qquad
\sigma_c\sigma_d=\sigma_{c+d}.
$$
These exhaust the $p=[L:K]$ automorphisms. This is an <Artin–Schreier extension>, and
$$
\boxed{\operatorname{Gal}(L/K)\cong(\mathbb F_p,+)\cong C_p}.
$$
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