= Solution
Use the standard <canonical anticommutation relations> for the two distinct <fermionic> orbitals. In particular $\{a,b^\dagger\}=\{a^\dagger,b\}=0$, in addition to the relations explicitly listed in the paper; these mixed relations are part of the usual two-orbital <fermionic Fock space> interpretation. Normalize the original <vacuum state> by $\langle0|0\rangle=1$ and fix the ordering
$$
|n_a n_b\rangle=(a^\dagger)^{n_a}(b^\dagger)^{n_b}|0\rangle.
$$
The <canonical anticommutation relations> give $a^2=(a^\dagger)^2=0$ and $aa^\dagger=1-a^\dagger a$. Consequently the <number operator> is an <orthogonal projector>:
$$
N_a^2=a^\dagger aa^\dagger a
=a^\dagger(1-a^\dagger a)a=N_a.
$$
The same holds for $N_b$. Hence each <eigenvalue> obeys $n^2=n$ and is either $0$ or $1$, expressing the <Pauli exclusion principle>. Also $[N_a,N_b]=0$, so the two <number operators> have a simultaneous <orthonormal basis> of <Fock states>:
$$
\begin{array}{c|c|c}
\text{state}&N_a\text{ eigenvalue}&N_b\text{ eigenvalue}\\\hline
|00\rangle=|0\rangle&0&0\\
|10\rangle=a^\dagger|0\rangle&1&0\\
|01\rangle=b^\dagger|0\rangle&0&1\\
|11\rangle=a^\dagger b^\dagger|0\rangle&1&1
\end{array}
$$
\b[Each <number operator> has <eigenvalues> $0,1$, each with <state degeneracy> two.] For $N_a$, the zero <eigenspace> is spanned by $|00\rangle,|01\rangle$ and the unit <eigenspace> by $|10\rangle,|11\rangle$; for $N_b$, these are respectively $|00\rangle,|10\rangle$ and $|01\rangle,|11\rangle$. Any normalized <linear combination> within one such <eigenspace> is also an <eigenstate> of that <number operator>.
The ordering convention will matter below: $b^\dagger a^\dagger|0\rangle=-|11\rangle$ and
$$
ab|11\rangle=aba^\dagger b^\dagger|0\rangle
=-aa^\dagger bb^\dagger|0\rangle=-|0\rangle.
$$
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