= Solution
An element lies in $\mathfrak p^*$ exactly when all its homogeneous components lie in $\mathfrak p$. Suppose $ab\in\mathfrak p^*$ but $a,b\notin\mathfrak p^*$. Choose the least-degree components $a_i\notin\mathfrak p$ and $b_j\notin\mathfrak p$. In the degree $i+j$ component of $ab$, every term other than $a_ib_j$ contains a lower component of $a$ or $b$ and hence lies in $\mathfrak p$. Since the whole component lies in $\mathfrak p$, it follows that $a_ib_j\in\mathfrak p$, contradicting primality. Thus
$$
\boxed{\mathfrak p^*\text{ is prime}.}
$$
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