[guided]We need two different features of $P$. The additive Cousin problem is controlled by holomorphic sheaf cohomology, so we need Steinness. The multiplicative Cousin problem is controlled by the exponential sequence, so we also need the topological vanishing $H^2(P,\mathbb{Z}_P)=0$.
Choose a center $a=(a_1,\ldots,a_n)\in \mathbb{C}^n$ and radii $r_1,\ldots,r_n\in (0,\infty]$ such that
\begin{align*}
P=\{z=(z_1,\ldots,z_n)\in \mathbb{C}^n: |z_j-a_j|<r_j \text{ for every } 1\leq j\leq n \}.
\end{align*}
When $r_j=\infty$, the $j$th coordinate condition is void. By the Steinness of Polydiscs, this polydisc is a Stein space.
Now we verify contractibility directly. Define
\begin{align*}
H: P\times [0,1] &\to P \\
(z,t) &\mapsto a+(1-t)(z-a).
\end{align*}
This map contracts every point linearly to the center $a$. To check that the contraction stays inside $P$, fix $z\in P$, $t\in [0,1]$, and an index $j$ with $r_j<\infty$. Then
\begin{align*}
|H_j(z,t)-a_j|=(1-t)|z_j-a_j|<r_j,
\end{align*}
because $0\leq 1-t\leq 1$ and $|z_j-a_j|<r_j$. If $r_j=\infty$, there is no coordinate restriction to check. Therefore $H$ is a well-defined continuous map $P\times[0,1]\to P$, with $H(z,0)=z$ and $H(z,1)=a$. This proves that $P$ is contractible.
The cohomology group used in the exponential sequence is sheaf cohomology with coefficients in the constant sheaf $\mathbb{Z}_P$. Since $P$ is an open subset of $\mathbb{R}^{2n}$, it is paracompact and locally contractible. The comparison theorem for constant sheaf cohomology and singular cohomology identifies $H^2(P,\mathbb{Z}_P)$ with singular cohomology $H^2_{\mathrm{sing}}(P,\mathbb{Z})$. Because $P$ is contractible, the homotopy invariance of singular cohomology gives
\begin{align*}
H^2(P,\mathbb{Z}_P)=0.
\end{align*}[/guided]