I think that you must mean $e$ to be the image of $\operatorname{diag}(-w, w, -w, w, \dotsc, -w, w)$, since otherwise $e$ does not lie in $\operatorname{GSp}_{2m}(K)$ when $m$ is odd and greater than $1$.
I am more comfortable dealing with the symplectic form with matrix $\begin{pmatrix} & w_0 \\ -w_0 \end{pmatrix}$, where $w_0 = \operatorname{antidiag}(1, \dotsc, 1)$. This is conjugate to your form by the permutation matrix corresponding to the element of $\operatorname{Sym}_{2m}$ given by $i \mapsto 2i - 1$ and $2m - i + 1 \mapsto 2i$ for $i \le m$. In particular, in these coördinates, your element $e$ (revised as I propose) is identified to the element $e = \operatorname{diag}(-w, \dotsc, -w, w, \dotsc, w)$ that you originally proposed, and the answer is as in your other question, with an obvious sign change: $\operatorname{GL}_m$ embeds in $\operatorname{Sp}_{2m}$ by $g \mapsto \begin{pmatrix} g \\ & \operatorname{Int}(w_0)g^{-\mathsf T} \end{pmatrix}$, and we may take the "2" to be represented by the image in $\operatorname{PGSp}_{2m}$ of $\begin{pmatrix} & I_m \\ -I_m \end{pmatrix}$.
Conjugating this back to your original set-up gives the representative $\operatorname{antidiag}(1, -1, 1, -1, \dotsc, 1, -1)$ (or possibly its opposite—I didn't check carefully, but it doesn't matter, since we're projecting to $\operatorname{PGSp}_{2m}$).