Example 5 If $k$ is a positive integer, and the two roots of the linear equation $(k-1) x^{2}-p x+k=0$ are both positive integers, then the value of $k^{p k}\left(p^{p}+k^{k}\right)+(p+k)$ is $\qquad$ .
Solution: From $x_{1} x_{2}=\frac{k}{k-1}$, we know $\frac{k}{k-1}$ is a positive integer.
$$
\begin{array}{l}
\therefore k-1=1, k=2 . \\
\text { So } x_{2}=2 \text {. } \\
\therefore x_{1}=1, x_{2}=2 \text {. }
\end{array}
$$

From $x_{1}+x_{2}=\frac{p}{k-1}$, we get $p=3$.
Therefore, the original expression $=2^{6}\left(3^{3}+2^{2}\right)+(3+2)=1989$.