= Solution
Insert the decorating-spin <observable> into the one bond that contains it. The <decorated-spin occupation insertion> uses the <matrix>:
$$
Q_{st}=\sum_{t_0=-1}^{1}t_0^2e^{Kt_0(s+t)}=2\cosh[K(s+t)]
=\begin{pmatrix}2\cosh2K&2\\2&2\cosh2K\end{pmatrix}_{st}.
$$
The <partition function> still contains $L$ decorating-spin sums; one of them has been replaced by this weighted sum. Hence
$$
\langle\tau_1^2\rangle_L=\frac{\operatorname{tr}(QV^{L-1})}{Z_L}
=\frac{q_+\lambda_+^{L-1}+q_-\lambda_-^{L-1}}{\lambda_+^L+\lambda_-^L},
\qquad q_+=2\cosh2K+2,\quad q_-=2\cosh2K-2.
$$
Taking the dominant <eigenvector> gives
$$
\boxed{\lim_{L\to\infty}\langle\tau_1^2\rangle_L=\frac{q_+}{\lambda_+}
=\frac{\cosh2K+1}{\cosh2K+2}=\frac{2\cosh^2K}{1+2\cosh^2K}.}
$$
This is the <probability> that the decorating site is in either nonzero state. It tends to $2/3$ at high <temperature>, when its three states are equally likely, and to one at low <temperature>, when the nonzero aligned states gain <Energy>[bond energy]. Unlike the <magnetization>, this even <observable> is not forced to vanish by <spin> reversal.
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