= Solution
Put $\beta=(k_BT)^{-1}$ and label the two <Spin>[spin values] in the order $+1,-1$. Assign each bond its <Boltzmann weight>,
$$
V_{st}=e^{-\beta E_0(s,t)},\qquad
V=\begin{pmatrix}a&b\\b&d\end{pmatrix},\qquad s,t\in\{+1,-1\}.
$$
The <symmetry> of the <Energy>[bond energy] makes $V$ a <Symmetric matrix>[real symmetric matrix]. For a fixed configuration the product of successive <matrix> entries is its <Boltzmann weight>. Summing over all <spins> and identifying the last index with the first therefore gives
$$
\boxed{Z_L=\sum_{\sigma_1,\ldots,\sigma_L}V_{\sigma_1\sigma_2}\cdots V_{\sigma_L\sigma_1}=\operatorname{tr}V^L=\lambda_+^L+\lambda_-^L,}
$$
where
$$
\lambda_\pm=\frac{a+d\pm\sqrt{(a-d)^2+4b^2}}2.
$$
This replaces a sum over $2^L$ configurations by the powers of a two-by-two <matrix>. For finite bond <energies> at $T>0$, every <matrix> entry is positive, so the <Perron–Frobenius theorem> gives $\lambda_+>0$ and $|\lambda_-|<\lambda_+$. Hence the <Helmholtz free energy> per <spin> tends to $-k_BT\log\lambda_+$. A diagonal local <observable> $O$ can be inserted at one site: $\langle O_1\rangle=\operatorname{tr}(OV^L)/Z_L$. Its limiting average is the <expectation> in the <Normalization>[normalized] <eigenvector> of $\lambda_+$.
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