Let me rename $p$ to $z$, because $p$ usually stands for prime numbers in the subject. I will assume that $z\geq 2$, but I will not assume that $z$ is an integer. The Dirichlet coefficients of $\zeta(s)^z$ form a generalized divisor function:
$$\zeta(s)^\nu=\sum_{n=1}^\infty\frac{\tau_z(n)}{n^s},\qquad \Re(s)>1.$$
The generalized divisor function is multiplicative, and on prime powers it is given by
$$\tau_z(p^\nu)=\binom{z+\nu-1}{\nu}.$$
Using this formula and its Taylor generating series $(1-x)^{-z}$, it is not hard to prove that
$$\tau_z(p^\nu)\leq\min\bigl((\nu+1)^z,z^\nu\bigr).$$
Then we can proceed as in the proof of Theorem 2 in Section I.5.2 of Tenenbaum's book "Introduction to analytic and probabilistic number theory" to see that, for any $t>1$,
$$\tau_z(n)\leq\exp\left(tz(2+\log\log n)+\frac{\log z\cdot\log n}{\log t}\right).$$
Choosing $t>1$ such that
$$t(\log t)^2=\frac{\log z\cdot\log n}{z(2+\log\log n)},\tag{1}$$
the above bound becomes
$$\tau_z(n)\leq\exp\left(\frac{\log z\cdot\log n}{(\log t)^2}+\frac{\log z\cdot\log n}{\log t}\right).$$
However, for the above choice of $t$,
$$\log t=\log\log n+O\bigl(\log z+\log\log\log n\bigr),\tag{2}$$
as can be seen by checking the cases $\log n\leq z$ and $\log n>z$ separately.
Let us now assume for simplicity that $\log n>z^2$. Then $\log n>t(\log t)^2\gg(\log n)^{1/2}$ by $(1)$, hence also $\log t\asymp\log\log n$. We conclude that
$$\tau_z(n)\leq n^{\frac{\log z}{\log\log n}\left(1+O\left(\frac{\log z+\log\log\log n}{\log\log n}\right)\right)},\qquad n>\exp(z^2).$$
The implied constants here are absolute, i.e., they do not depend on $z$.

**Remark.** One can cover the range $1\leq z<2$ similarly, but with more care. In particular, everthing before $(2)$ is valid in this range, but $(2)$ fails e.g. when $n$ is fixed and $z$ approaches $1$.