Let's say I have a modular form $f \in M_{k}^{!}(\Gamma_{0}(N))$ which is holomorphic at the cusps except $\infty$, and I apply the Hecke operator $T_{k}(p)$ at a prime $p$ not dividing $N$ to get the summation
$$p^{\frac{k}{2} -1}\left( \sum_{j=0}^{p-1} f \mid_{k} \begin{pmatrix} 1 & j \\ 0 & p \end{pmatrix} + f \mid_{k} \begin{pmatrix} p & 0 \\ 0 & 1 \end{pmatrix} \right).$$
I want to then say this new summation is holomorphic at the cusps. There's a trick I saw in a paper: let $r \in \mathbb{Q}$ be a cusp of $\Gamma_{0}(N)$ that is not equivalent to $\infty$, and let $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_{2}(\mathbb{Z}) \setminus \Gamma_{0}(N)$ with $\gamma \infty = r$. A "standard argument" shows that
$$\begin{pmatrix} 1 & j \\ 0 & p \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \frac{a + cj}{\lambda} & * \\ \frac{cp}{\lambda} & * \end{pmatrix} \begin{pmatrix} \lambda & * \\ 0 & \frac{p}{\lambda} \end{pmatrix}$$
(which I believe to be true, so I'll take that for granted for now). The first matrix on the right is in the full modular group and not in $\Gamma_{0}(N)$. The author of the paper then claims that each summand on $j$ above is now holomorphic at the cusps. Why is this?
[The same trick is used for the last summand $f \mid_{k} \begin{pmatrix} p & 0 \\ 0 & 1 \end{pmatrix}$.]