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I don't know if this is always true, and I don't think this is known in all cases, even after inverting $p$.

What is known and what is more or less formal to prove (it depends of your starting point, and in particular, of how you define the "$\Lambda$-adic Hecke algebra of tame level $N$) is that there is a natural surjection, whose kernel is nilpotent, from ${\mathbb T}/p_\kappa {\mathbb T}$ to the Hecke algebra of the space of overconvergent modular forms of weight $\kappa$ and slope 0. But then, the question of whether the kernel is trivial is difficult to settle without knowing more on the local structure of $\mathbb T$ near the points above $\kappa$.

Edit to answer HP's comment below : Let us consider a "toy model": imagine your base ring is $\mathbb{Z}_p$, your module of ordinary p-adic form is $M=\mathbb{Z}_p^2$, and there is only one Hecke operator around, namely $T$ acting on the canonical basis of $M$ by $T(e_1)=p e_1$, $T(e_2)=0$. Then $\mathbb{T}=\mathbb{Z}_p[T]/T(T-p)$ is free of rank $2$ over $\mathbb{Z}_p$, but on $M/pM$, $T$ acts like $0$ et the Hecke algebra on $M/pM$ is just $\mathbb{F}_p$ which is not $\mathbb{T}/p$.

Now what I say is not that this kind of examples actually happen on the eigencurve, but that without supplementary hypothesis (like level $1$, artfully restricting to new forms, etc.) no one, as far as I know, can rule this out.

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I don't know if this is always true, and I don't think this is known in all cases, even after inverting $p$.

What is known and what is more or less formal to prove (it depends of your starting point, and in particular, of how you define the "$\Lambda$-adic Hecke algebra of tame level $N$) is that there is a natural surjection, whose kernel is nilpotent, from ${\mathbb T}/p_\kappa {\mathbb T}$ to the Hecke algebra of the space of overconvergent modular forms of weight $\kappa$ and slope 0. But then, the question of whether the kernel is trivial is difficult to settle without knowing more on the local structure of $\mathbb T$ near the points above $\kappa$.