If we view permutation as runs of red balls interspaced with runs of blue balls, then the requirement is that the marked ball is at the even position within its run.
Let $t$ be the number of red runs; $r_i$ and $b_i$ be the number of red and blue runs of length $i$, respectively.
$$(n-k)!(k-1)!\sum_{t\geq 0} \sum_{1r_1 + 2r_2 + \dots = k\atop r_1 + \dots + r_k = t} \binom{t}{r_1,\dots,r_n} (r_2 + r_3 + 2(r_4+r_5) + \dots)$$ $$\bigg(2\sum_{1b_1 + 2b_2 + \dots = n-k\atop b_1 + \dots + b_k = t} \binom{t}{b_1,\dots,b_n} +\sum_{1b_1 + 2b_2 + \dots = n-k\atop b_1 + \dots + b_k = t-1} \binom{t-1}{b_1,\dots,b_n} +\sum_{1b_1 + 2b_2 + \dots = n-k\atop b_1 + \dots + b_k = t+1} \binom{t+1}{b_1,\dots,b_n} \bigg)$$ $$=(n-k)!(k-1)!\sum_{t\geq 0} \sum_{1r_1 + 2r_2 + \dots = k\atop r_1 + \dots + r_k = t} \binom{t}{r_1,\dots,r_n} (r_2 + r_3 + 2(r_4+r_5) + \dots)$$ $$\bigg(2\binom{n-k-1}{t-1} + \binom{n-k-1}{t-2} + \binom{n-k-1}{t} \bigg)$$ $$=(n-k)!(k-1)!\sum_{t\geq 0} \sum_{1r_1 + 2r_2 + \dots = k\atop r_1 + \dots + r_k = t} \binom{t}{r_1,\dots,r_n} (r_2 + r_3 + 2(r_4+r_5) + \dots)\binom{n-k+1}{t}$$ $$=(n-k)!(n-k+1)!\frac1{k}\sum_{t\geq 0} \frac1{(n-k+1-t)!}\sum_{1r_1 + 2r_2 + \dots = k\atop r_1 + \dots + r_k = t} \frac{k!}{r_1!\cdots r_n!} (r_2 + r_3 + 2(r_4+r_5) + \dots)$$
In terms of Bell polynomials this can be written as $$=(n-k)!(n-k+1)!\frac1{k}\frac{\partial}{\partial x}\left.\sum_{t\geq 0} \frac1{(n-k+1-t)!} B_{k}(1!,2!x,3!x,4!x^2,5!x^2,\dots)\right|_{x=1}$$ Then using the generating function for Bell polynomials we have
\begin{split} &\left.\frac{\partial}{\partial x}\sum_{t\geq 0} \frac1{(n-k+1-t)!} B_{k}(1!,2!x,3!x,4!x^2,5!x^2,\dots)\right|_{x=1} \\ &= \left.\frac{\partial}{\partial x} [y^{n-k+1}t^k]\ \exp(y) \exp(y (t + xt^2 + xt^3 + x^2t^4 + x^2t^5 +\dots))\right|_{x=1} \\ &= [t^k] \left.\frac{\partial}{\partial x}(1 + t + xt^2 + xt^3 + x^2t^4 + x^2t^5 +\dots)^{n-k+1}\right|_{x=1} \\ &=[t^k]\ (n-k+1) (1+ t + t^2 + t^3 + \dots)^{n-k} (t^2 + t^3 + 2t^4 + 2t^5 + \dots) \\ & = (n-k+1) [t^k] \frac{t^2}{(1-t)^{n-k+2}(1+t)} \end{split}
and the rest is trivial.