= Solution
First, $\sqrt{\mathfrak q^*}=\mathfrak p$. Indeed, each homogeneous element of $\mathfrak p=\sqrt{\mathfrak q}$ has a power in $\mathfrak q$, and that power is homogeneous and hence lies in $\mathfrak q^*$; finite homogeneous generators of the Noetherian ideal $\mathfrak p$ give the assertion for every element.
Now suppose $ab\in\mathfrak q^*$ and $a\notin\mathfrak p$. Choose the least homogeneous component $a_i\notin\mathfrak p$. If $b\notin\mathfrak q^*$, choose the least component $b_j\notin\mathfrak q$. The degree $i+j$ component of $ab$ differs from $a_ib_j$ by terms in $\mathfrak q$. Since it lies in $\mathfrak q$, we get $a_ib_j\in\mathfrak q$. The $\mathfrak p$-primary property and $a_i\notin\mathfrak p$ imply $b_j\in\mathfrak q$, a contradiction. Hence $b\in\mathfrak q^*$, proving that
$$
\boxed{\mathfrak q^*\text{ is }\mathfrak p\text{-primary}.}
$$
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