Let $S\_k$ be the space of weight 12 cusp forms of $\Gamma\_0(p)$, ($p$ prime), then Sage tells that $\dim S\_k^{\text{new}}=\dim S\_k-2$. Thus the old forms spans a 2-dimensional subspace. One of the obvious eigenbase vectors is $\Delta(z)=q\prod\_{n=1}^\infty(1-q^n)^{24}$. What does the other eigenbase vector look like? 



Weight 12 cusp forms for $\Gamma_0(p)$
Per suggestion of GH from MO in the comments, I'm turning my comment into an answer.


In general, the space $S\_{k}(\Gamma\_0(N))$ decomposes as $S\_{k}^{\text{new}}(\Gamma\_0(N)) \oplus S\_{k}^{\text{old}}(\Gamma\_0(N))$. Here the space of oldforms is spanned by the forms $f(dz)$, where $f$ is a newform of weight $k$ and level $\Gamma\_0(N')$, with $N' < N$ and $(N'\cdot d) \mid N$. The space $S^{\text{new}}$ is the orthogonal complement of $S^{\text{old}}$ under the Petersson inner product. Moreover, for a newform $f$ at level $N'$, the forms $f(dz)$ and $f(d'z)$ have the same Hecke eigenvalues (away from primes dividing $N$).


Now we apply this to the situation at hand, where $k = 12$ and $N = p$ is prime. There is only one form of weight $12$ and level $1$, which is $\Delta$ (up to scalar multiple). Since $p$ is prime, all oldforms must arise from forms of level $1$, so the space of oldforms has dimension $2$: it is spanned by $\Delta(z)$ and $\Delta(pz)$, and both have the same Hecke eigenvalues away form $p$.


