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You can trivialize that bundle over the same open sets on which $\pi$ is trivial with essentially the same trivialization---up to composing with $\rho$.

Later: The easiest way to see this is, I think the following. If your initial principal $G$-bundle is trivial, then it is more or less evident that $P\times_\rho\mathbb C^n$ is also trivial. Now if $\pi$ is not trivial but its restriction $\pi|U:\pi^{-1}(U)\to \pi|_U:\pi^{-1}(U)\to U$ is trivial over an open subset $U\subseteq B$, then by the previous observation and a little unravelling of the notation, we have that $P\times\rho\mathbb P\times_\rho\mathbb C^n$ is trivial over $U$.

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You can trivialize that bundle over the same open sets on which $\pi$ is trivial with essentially the same trivialization---up to composing with $\rho$.

Later: The easiest way to see this is, I think the following. If your initial principal $G$-bundle is trivial, then it is more or less evident that $P\times_\rho\mathbb C^n$ is also trivial. Now if $\pi$ is not trivial but its restriction $\pi|U:\pi^{-1}(U)\to U$ is trivial over an open subset $U\subseteq B$, then by the previous observation and a little unravelling of the notation, we have that $P\times\rho\mathbb C^n$ is trivial over $U$.

4 undid change, since it didn't work (there is no problem in preview)

You can trivialize that bundle over the same open sets on which $\pi$ is trivial with essentially the same trivialization---up to composing with $\rho$.

Later: The easiest way to see this is, I think the following. If your initial principal $G$-bundle is trivial, then it is more or less evident that $P\times_\rho\mathbb C^n$ is also trivial. Now if $\pi$ is not trivial but its restriction $\pi|U:\pi^{-1}(U)\to U$ is trivial over an open subset $U\subseteq B$, then by the previous observation and a little unravelling of the notation, we have that $P\times\rho\mathbb C^n$ is trivial over $U$.

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