= Solution
The two <protons> are identical spin-one-half <fermions>, so the total state is antisymmetric under exchange. In the electronic and vibrational <ground state>, an even rotational $J$ has a symmetric spatial wavefunction and therefore uses the antisymmetric nuclear-spin <singlet state>; odd $J$ uses the symmetric <triplet state>. Thus <parahydrogen> has one nuclear-spin state, and <orthohydrogen> has three. Each rotational level also has spatial <state degeneracy> $2J+1$.
The <translational partition functions> and all common electronic/vibrational factors cancel in the ratio of equilibrium populations. Using the same zero of <energy> for the two <spin> species gives
$$
\boxed{K=\frac{n_o}{n_p}=\frac{q_o}{q_p}
=\frac{3\sum_{J\ {\rm odd}}(2J+1)e^{-J(J+1)\theta_{\rm rot}/T}}
{\sum_{J\ {\rm even}}(2J+1)e^{-J(J+1)\theta_{\rm rot}/T}}.}
$$
Subtracting the $J=1$ <energy> separately from all odd levels without restoring that offset in the equilibrium ratio would incorrectly remove the low-temperature preference for <parahydrogen>. The <nuclear spin statistics> account for the factor three, independently of the spatial factor $2J+1$.
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