I think I was over thinking it. It doesn't seem that difficult, but let me illustrate with an example.
Given we are interested in 4 species of particles, we would expect a total Hamiltonian like,
\begin{align}
\mathcal{H}\big(\mathbb{Q}, \mathbb{P}; \vec{N}\big) &= \sum_{i=1}^4 \mathcal{H}_i\big(\mathbb{Q}_i \subset \mathbb{Q}, \mathbb{P}_i \subset \mathbb{P} ; \vec{N}_i\big) + U\big(\mathbb{Q}, \mathbb{P}\big)
\end{align}
where $ | \mathbb{Q}_i | = | \mathbb{P}_i | = N_i $.
Thinking about a single species system, computing the Isothermal-Isobaric ensemble gets you the Gibbs free energy,
\begin{aligned}
G(T, P, N) &= -k T\log \Delta (N,P,T) \\
&= -k T \log \left[ {\frac {C}{h^{3}}}\int {\mathrm {e}}^{{-\beta (H(q,p) + PV)}}~d^{{3N}}q~d^{{3N}}p~dV \right]
\end{aligned}
If you compute this for a multi-species particle, then you can get $ \sum_i \mu_i \nu_i = \sum_i \frac{dG}{dN_i}\frac{dN_i}{d\xi} = 0 $ as the chemical equilibrium point as a function of $ P, T $.
Ideal Gas Example
For an ideal gas (see Section 2.4 these lecture notes),
\begin{align}
G(T, P, N) = -N k_B T \log \left[ \frac{k_B T}{P} \cdot \frac{(2\pi m k_B T)^{3N/2}}{h^{3N}} \right]
\end{align}
in chemistry, its written like $ G(T, P, N) = G(P_0) + nRT \ln\left(\frac{P}{P_0}\right) $. I'm ignoring the fact that this differs from the Stat Mech version and just using the stat mech version.
Going back to the multi-species problem, if we drop the interaction term $ U\big(\mathbb{Q}, \mathbb{P}\big) $ and assume an ideal gas, then
\begin{align}
G(T, P, N) &= -\sum_i N_i k_B T \log \left[ \frac{k_B T}{P} \cdot \frac{(2\pi m_i k_B T)^{3N_i/2}}{h^{3N_i}} \right] \\
&= \sum_i G_i
\end{align}
I am just taking $ P, T $ to be the same for each species because they are Lagrange multipliers and Laplace transform variables, so having multiple doesn't make much sense to me
Then,
\begin{align}
\sum_i \mu_i \nu_i = \sum_i \frac{dG_i}{dN_i}\frac{dN_i}{d\xi}
= - \sum_i \nu_i k_B T \log \left[ \frac{k_B T}{P} \right] + 2 \big( N^{(0)}_i + \nu_i \xi \big) k_B T \log \left[\frac{(2\pi m_i k_B T)^{3/2}}{h^{3}} \right]
\end{align}
Setting this equal to zero, taking $ m_i = m $, and the translation partition function $ {\displaystyle \Lambda ={\frac {h}{\sqrt {2\pi mk_{B}T}}}} $,
\begin{align}
\xi = \frac{\sum_i \log \left[ \left(\frac{k_B T}{P}\right)^{\nu_i} \Lambda^{6 N^{(0)}_i} \right]}{\sum_i \log \left[\Lambda^{-6 \nu_i } \right]}
\end{align}
