Only one saturated step can lie over a fixed contracted prime in R[x]

Passing to the quotient by $\mathfrak{p}R[x]$ and then localizing away from the nonzero elements of $R\mathrm{/}\mathfrak{p}$, [L1] and [L2] identify the fiber over $\mathfrak{p}$ with the prime spectrum of $K[x]$. Under this identification, the images of $P$ and $Q$ are comparable prime ideals of $K[x]$, with the image of $P$ properly contained in the image of $Q$.

By [L3], the only way two comparable primes in $K[x]$ can be strictly nested is for the smaller one to be $(0)$ and the larger one to be a nonzero maximal prime. Therefore the image of $P$ in $K[x]$ is zero. Contracting back through [L2], this means $P=\mathfrak{p}R[x]$. The same description also shows that no third prime can lie strictly between $P$ and $Q$, because no third prime lies strictly between $(0)$ and a nonzero prime in the PID $K[x]$.

Hence over a fixed contracted prime in $R[x]$ there is at most one extra strict step, and the lower prime in such a pair is exactly the extended prime $\mathfrak{p}R[x]$.