= Solution
The <modular discriminant> has the convergent product
$$
\Delta(\tau)=q\prod_{n\geq1}(1-q^n)^{24},\qquad q=e^{2\pi i\tau},
$$
is a weight-twelve level-one <cusp form>, and has order one at infinity. Applying the <degeneracy map for cusp forms> with $M=1,N=D=p$ gives $\boxed{\Delta(p\tau)\in S_{12}(\Gamma_0(p))}$. The original form $\Delta$ is also a <cusp form> on this subgroup.
There are precisely two <modular cusps> for prime level, represented by infinity and zero. To see this directly, write a rational cusp as $a/c$ in lowest terms. If $p\mid c$, completing $(a,c)^T$ to a determinant-one matrix gives a member of $\Gamma_0(p)$ taking infinity there. If $p\nmid c$, the equation $Ac-aC=1$ has a solution with $p\mid C$: adjust a Bézout solution modulo $p$, using the invertibility of $c$. The matrix $\begin{pmatrix}A&a\\C&c\end{pmatrix}$ then lies in $\Gamma_0(p)$ and takes zero there. The two classes are distinct because their denominator divisibility is preserved by the group.
At infinity the <cusp width> is one, so $q$ is the local parameter. The products start with $q$ and $q^p$, giving orders one and $p$. At zero use $S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. Since $ST^hS^{-1}=\begin{pmatrix}1&0\\-h&1\end{pmatrix}$, the <cusp width> is $p$ and the parameter is $q_0=e^{2\pi i\tau/p}$. The weight-twelve inversion law of the <modular discriminant> gives
$$
\Delta[ S]_{12}(\tau)=\Delta(\tau),\qquad
\big(\Delta(p\tau)\big)[ S]_{12}=\tau^{-12}\Delta(-p/\tau)=p^{-12}\Delta(\tau/p).
$$
Here determinant one makes the printed and ordinary <slash operators for modular forms> agree. In terms of $q_0$, the leading terms are $q_0^p$ and $p^{-12}q_0$. Hence the <prime-level cusp orders of discriminant degeneracy forms> are
$$
\boxed{\begin{array}{c|cc}&\Delta(\tau)&\Delta(p\tau)\\\hline\infty&1&p\\0&p&1\end{array}.}
$$
The scalar $p^{-12}$ changes no order of vanishing.
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