In 1960, Reed and Solomon suggest the codeword for a message $[x_0\ x_1\ \ldots\ x_k]$ as follows:
$$[P_{(0)}\ P_{(\alpha)}\ P_{(\alpha^2)}\ \cdots\ P_{(\alpha^{2^m-1})}]$$
Where
$$P_{(t)}=x_0 +x_1t+x_2t^2+\cdots +x_{k-1}t^{k-1}$$ How is this equivalent to the BCH code generated by the following polynomial?
$$g{(x)} = \prod_{i=1}^{n-k} x-\alpha^i$$
Your assertion is not quite correct: $[P_{(0)}\ P_{(\alpha)}\ P_{(\alpha^2)}\ \cdots\ P_{(\alpha^{2^m-1})}]$ is a codeword in an extended cyclic (Reed-Solomon) code which is also an extended cyclic BCH code. The $P_{(0)}$ symbol is an overall parity check on the remaining symbols which do, in fact, form a cyclic BCH code in the sense that the corresponding codeword polynomials are multiples of the generator polynomial of a cyclic code which has $n-k$ consecutive powers of $\alpha$ as its roots.
Specifically, if $P(x)$ is a polynomial of degree $k-1$ or less with coefficients in $\mathbb F_{2^m}$ and $\alpha$ is a primitive element of this field, then $[P(1)\ P(\alpha)\ P(\alpha^2)\ \cdots\ P(\alpha^{2^m-2})]$ is a codeword in a cyclic $[2^m-1,k]$ Reed-Solomon code. The corresponding codeword polynomial is $$\hat{P}(z) = P(1) + P(\alpha)z + P(\alpha^2)z^2 + \cdots P\left(\alpha^{2^m-2}\right)z^{2^m-2}.$$ Think of $[P(1)\ P(\alpha)\ P(\alpha^2)\ \cdots\ P(\alpha^{2^m-2})]$ as the finite-field discrete Fourier transform of the vector $[P_0\ P_1\ P_2\ \cdots \ P_{k-1}\ 0 \ 0 \ \cdots \ 0]$. The inverse discrete Fourier transform says that $$P_i = \hat{P}\left(\alpha^{-i}\right) = \hat{P}\left(\alpha^{2^m-1-i}\right)$$ and since we know that $P_{2^m-2}, P_{2^m-2}, \cdots, P_{k+1}, P_k$ all have value $0$, we deduce that $$\hat{P}(\alpha) = \hat{P}(\alpha^2) = \cdots = \hat{P}\left(\alpha^{2^m-1-k}\right) = 0,$$ that is, $\hat{P}(z)$ is a multiple of $$g(z) = \sum_{i=1}^{2^m-1-k}(z-\alpha^i).$$