Let $\mathcal{P}=\{p_1,p_2,\ldots,p_k\}$. Then the generating function of $f(n)$ is
\begin{align}
G_{f}(t):=\sum_{n\ge 0}f(n)e^{-nt}=\prod_{\ell_1,\ell_2,\ldots,\ell_k\ge 0}\frac{1}{1-\exp\left(-t\prod_{1\le j\le k}p_j^{\ell_j}\right)},
\end{align}
where $t>0$. This means
$$f(n)\le G_f(t)e^{nt}$$
for any integer $n>0$ and any real number $t>0$. On the other hand, using part integration,
\begin{align*}
\log G_{f}(t)&=-\sum_{\ell_1,\ell_2,\ldots,\ell_k\ge 0}\log\left(1-\exp\left(-t\prod_{1\le j\le k}p_j^{\ell_j}\right)\right)\\
&=-\int_{1-}^{\infty}\log\left(1-\exp\left(-t x\right)\right)\,dS(x)\\
&=\int_{1}^{\infty}\frac{tS(x)}{e^{xt}-1}\,dx,
\end{align*}
where
$$S(x)=\sum_{\substack{\prod_{1\le j\le k}p_j^{\ell_j}\le x,\;\ell_1,\ell_2,\ldots,\ell_k\ge 0}}1.$$
Clearly,
$$S(x)\le \prod_{1\le j\le k}\left(1+\left\lfloor\frac{\log x}{\log p_1}\right\rfloor\right)=\frac{(\log x)^k}{\prod_{1\le j\le k}\log p_j}+O\left((\log x)^{k-1}\right).$$
Thus one can from above obtain
\begin{align*}
\log G_{f}(t)
&\le \frac{\log^{k+1}(1/t)}{(k+1)\prod_{1\le j\le k}\log p_j}+O(\log^k(1/t)).
\end{align*}
Taking $t=1/n$ then there exist a constant $C_k>0$ such that
$$f(n)\le G_f(t)e^{nt}\le \exp\left(1+\frac{\log^{k+1}n}{(k+1)\prod_{1\le j\le k}\log p_j}+O(\log^kn)\right)\le C_k e^{\log^{k+1}n}$$
by note that $(k+1)\prod_{1\le j\le k}\log p_j>1$.