= Solution
Pass to the graded domain $R=S/\mathfrak p^*$ and let $\overline{\mathfrak p}=\mathfrak p/\mathfrak p^*$. This is a nonzero prime containing no nonzero homogeneous element. Localize at the multiplicative set $U$ of all nonzero homogeneous elements. Every nonzero homogeneous element of $U^{-1}R$ is a unit; its nonzero graded pieces are one-dimensional over the degree-zero field, so after reindexing degrees this localization is a Laurent polynomial ring $K[t,t^{-1}]$. The extended prime $U^{-1}\overline{\mathfrak p}$ is therefore a nonzero prime of height one.
Any prime strictly between $0$ and $\overline{\mathfrak p}$ would remain a nonzero prime strictly below it after localization, impossible in $K[t,t^{-1}]$. Contracting back proves that no prime lies strictly between $\mathfrak p^*$ and $\mathfrak p$. The graded height theorem, equivalently the same localization argument applied to saturated chains, then gives
$$
\boxed{\operatorname{ht}(\mathfrak p)
=\operatorname{ht}(\mathfrak p^*)+1.}
$$
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