[guided]The stalk of $\mathfrak{m}_p$ at $p$ is the ordinary maximal ideal of the local analytic ring. We denote this maximal ideal by
\begin{align*}
\mathfrak{m}_{X,p} \subset \mathcal{O}_{X,p}.
\end{align*}
Its elements are precisely the germs at $p$ of holomorphic functions that vanish at $p$. Thus
\begin{align*}
(\mathfrak{m}_p)_p=\mathfrak{m}_{X,p}.
\end{align*}
The previous step therefore says that the germs $(g_1)_p,\ldots,(g_N)_p$ generate the maximal ideal $\mathfrak{m}_{X,p}$ over the local ring $\mathcal{O}_{X,p}$.
To extract first-order information, define
\begin{align*}
V := \mathfrak{m}_{X,p}/\mathfrak{m}_{X,p}^2.
\end{align*}
This quotient keeps the linear part of a germ and kills all terms of order at least two. For each $j \in \{1,\ldots,N\}$, define
\begin{align*}
\overline{g}_j := (g_j)_p + \mathfrak{m}_{X,p}^2 \in V.
\end{align*}
Because every element of $\mathfrak{m}_{X,p}$ is an $\mathcal{O}_{X,p}$-linear combination of the germs $(g_j)_p$, reducing modulo $\mathfrak{m}_{X,p}^2$ shows that every class in $V$ is a complex-linear combination of $\overline{g}_1,\ldots,\overline{g}_N$. Indeed, if
\begin{align*}
h_p=\sum_{j=1}^N a_j (g_j)_p
\end{align*}
with $a_j \in \mathcal{O}_{X,p}$, then modulo $\mathfrak{m}_{X,p}^2$ only the scalar values $a_j(p)$ contribute:
\begin{align*}
h_p+\mathfrak{m}_{X,p}^2
=
\sum_{j=1}^N a_j(p)\,\overline{g}_j.
\end{align*}
The terms $(a_j-a_j(p))(g_j)_p$ lie in $\mathfrak{m}_{X,p}^2$ because $a_j-a_j(p)\in \mathfrak{m}_{X,p}$ and $(g_j)_p\in \mathfrak{m}_{X,p}$.
Finally, we identify this quotient with the cotangent space. The standard cotangent-space isomorphism is
\begin{align*}
\delta_p: \mathfrak{m}_{X,p}/\mathfrak{m}_{X,p}^2 &\to T_p^*X,\\
h_p+\mathfrak{m}_{X,p}^2 &\mapsto dh(p).
\end{align*}
It is well-defined because a germ with zero first-order part has differential zero at $p$, and every cotangent vector is the differential of a local holomorphic coordinate combination. Since $X$ has complex dimension $n$, the cotangent space $T_p^*X$ has complex dimension $n$. Hence $V$ also has complex dimension $n$.[/guided]