-1 -

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14 (1) Since
$$
a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{p^{2}}{a_{n}}\right), b_{n}=\frac{a_{n}+p}{a_{n}-p}\left(n \in \mathbf{N}^{*}, p>0\right)
$$

Therefore,
$$
\begin{aligned}
b_{n+1} & =\frac{a_{n+1}+p}{a_{n+1}-p}=\frac{a_{n}+\frac{p^{2}}{a_{n}}+2 p}{a_{n}+\frac{p^{2}}{a_{n}}-2 p}=\frac{\left(a_{n}+p\right)^{2}}{\left(a_{n}-p\right)^{2}} \\
& =b_{n}^{2}>0\left(n \in \mathbf{N}^{*}, p>0\right) .
\end{aligned}
$$

Thus, $\lg b_{n+1}=2 \lg b_{n}$. Since $p>0$, we have $b_{n}=\frac{a_{n}+p}{a_{n}-p} \neq 1$, $\lg b_{n} \neq 0$, so $\frac{\lg b_{n+1}}{\lg b_{n}}=2$.

Therefore, the sequence $\left\{\lg b_{n}\right\}$ is a geometric sequence with a common ratio of 2 and the first term $\lg b_{1}=\lg 3$. Hence, $\lg b_{n}=2^{n-1} \cdot \lg 3$, and $b_{n}=3^{2^{n-1}}$.
(2) From $b_{n}=\frac{a_{n}+p}{a_{n}-p}\left(n \in \mathbf{N}^{*}, p>0\right)$, we get
$$
a_{n}=\frac{b_{n}+1}{b_{n}-1} \cdot p=p+\frac{2 p}{3^{2^{n-1}}-1},
$$

Therefore,
$$
\frac{a_{n}-p}{a_{n+1}-p}=\frac{\frac{2 p}{3^{2^{n-1}}-1}}{\frac{2 p}{3^{2^{n}}-1}}=\frac{3^{2^{n}}-1}{3^{2^{n-1}}-1}=\frac{\left(3^{2^{n-1}}\right)^{2}-1}{3^{2^{n-1}}-1}=3^{2^{n-1}}+1 .
$$

The proposition is proved.
(3) When $n \geqslant 2$,
$$
a_{n+1}-p=\frac{a_{n}-p}{3^{2^{n-1}}+1} \leqslant \frac{1}{10}\left(a_{n}-p\right),
$$

Thus, when $n=2$,
$$
\begin{array}{c}
S_{n}=S_{2}=a_{1}+a_{2}=2 p+\frac{5}{4} p=\frac{13}{4} p, \\
\left(n+\frac{23}{18}\right) p=\left(2+\frac{23}{18}\right) p=\frac{59}{18} p,
\end{array}
$$

And
$$
\frac{13}{4} p-\frac{59}{18} p=-\frac{1}{36} p<0,
$$

That is, $S_{n}<\left(n+\frac{23}{18}\right) p$.
When $n \geqslant 3$, $a_{3}-p \leqslant \frac{1}{10}\left(a_{2}-p\right), a_{4}-p \leqslant \frac{1}{10}\left(a_{3}-p\right), \cdots$, $a_{n}-p \leqslant \frac{1}{10}\left(a_{n-1}-p\right)$, so $a_{1}=2 p, a_{2}=\frac{5 p}{4}$, thus
$$
10 S_{n}-\frac{65}{2} p-10(n-2) p \leqslant S_{n}-a_{n}-2 p-(n-2) p,
$$

Therefore,
$$
S_{n} \leqslant\left[(n-2)+\frac{61}{18}-\frac{3^{2^{n-1}}+1}{9\left(3^{2^{n-1}}-1\right)}\right] \cdot p<\left(n+\frac{23}{18}\right) p .
$$

In summary, when $n \geqslant 2$, $S_{n}<\left(n+\frac{23}{18}\right) p$.