We have to move all operators that annihilate the reference state to the right of those that create on the reference state. Thus,
$$ \begin{align} \sum_{pq}\varepsilon_q^p a^\dagger_p a_q &=\sum_{ab}\varepsilon_b^a a^\dagger_a a_b +\sum_{ai}\varepsilon_i^a a^\dagger_a a_i +\sum_{ai}\varepsilon_a^i a^\dagger_i a_a +\sum_{ij}\varepsilon_j^i a^\dagger_i a_j \tag{6}\\ &=\sum_{ab}\varepsilon_b^a a^\dagger_a a_b +\sum_{ai}\varepsilon_i^a a^\dagger_a a_i +\sum_{ai}\varepsilon_a^i a^\dagger_i a_a +\sum_{ij}\varepsilon_j^i \left(-a_ja^\dagger_i +\delta_i^j\right) \tag{7}\\ &=\sum_{ab}\varepsilon_b^a a^\dagger_a a_b +\sum_{ai}\varepsilon_i^a a^\dagger_a a_i +\sum_{ai}\varepsilon_a^i a^\dagger_i a_a -\sum_{ij}\varepsilon_j^i a_ja^\dagger_i +\sum_i \varepsilon_i^i \tag{8}\\ &=\sum_{pq}\varepsilon_q^p \left\{a^\dagger_p a_q\right\} +\sum_i \varepsilon_i^i \tag{9} \end{align} $$