Let r be a root of q(x). Evaluate q(r+j) for any integer j. By the frobenius homomorphism, the result is rp+jp-r-j-a, which is rp-r-a, or 0. If r is a root then every r+j is a root. That's p distinct roots, and there can be no more, hence q(x) splits.
Conversely suppose q(x) has a proper irreducible factor s(x), but has no roots in K. Let s have the root r, which lies in a field extension E/K. Within E, s has the roots r+j for every j, and these are distinct, so s has degree p, s divides into q, and s = q. This is a contradiction, hence q splits in K. Either q is irreducible or it splits.