= Solution
<Chemical equilibrium> for $\mathrm M\rightleftharpoons\mathrm{M^+}+\mathrm{e^-}$ requires
$$
\mu_{\rm M}=\mu_{\rm M^+}+\mu_e.
$$
Insert the given <chemical potential> form and define the positive <ionization energy>
$$
E_I=\varepsilon_{\rm M^+}+\varepsilon_e-\varepsilon_{\rm M}.
$$
After collecting logarithms,
$$
k_{\rm B}T\ln\left(\frac{f_{\rm M}c_{\rm M^+}c_e}
{c_{\rm M}f_{\rm M^+}f_e}\right)+E_I=0,
$$
so
$$
\frac{c_{\rm M^+}c_e}{c_{\rm M}}
=\frac{f_{\rm M^+}f_e}{f_{\rm M}}e^{-E_I/k_{\rm B}T}.
$$
Define a dimensionless concentration <equilibrium constant> using a standard <number density> $c^\circ$:
$$
\boxed{K_c=\frac{(c_{\rm M^+}/c^\circ)(c_e/c^\circ)}{c_{\rm M}/c^\circ}
=\frac{f_ef_{\rm M^+}}{f_{\rm M}}\frac1{c^\circ}e^{-E_I/k_{\rm B}T}.}
$$
Each $f_i$ is a <molecular partition function density>, with units of inverse volume. Consequently $f_ef_{\rm M^+}/f_{\rm M}$ has density units and division by $c^\circ$ makes $K_c$ dimensionless. All ground-state energies must use the same <energy> reference.
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