Assuming $\zeta$ is a primitive $p$th root of unity, yes. We can assume $p \
eq 2$ since the result is obvious for $X^2 - 2$ over $\mathbb{Q}$. Then the degree of $\mathbb{Q}(\zeta)$ over $\mathbb{Q}$ is $p-1$, and in particular is coprime to $p$.
Since $X^p - p$ is irreducible over $\mathbb{Q}$ (Eisenstein), the degree of $\mathbb{Q}(\sqrt[p]{p})$ over $\mathbb{Q}$ is $p$. It follows from the multiplicativity of degrees in extensions that the degree of $\mathbb{Q}(\zeta, \sqrt[p]{p}$) over $\mathbb{Q}(\zeta)$ must be $p$.