Take $\mathbb{R}P^{n-1}\hookrightarrow\mathbb{R}P^{n+k}$ as embedding on the last $n$ coordinates and take a point $l=[x_0:\ldots : x_k:y_1:\ldots :y_n]\in\mathbb{R}P^{n+k}-\mathbb{R}P^{n-1}$.
We define a map $\pi: \mathbb{R}P^{n+k}-\mathbb{R}P^{n-1}\hookrightarrow\mathbb{R}P^k$ by sending $l$ to $[x_1:\ldots :x_k]$. So $l$ is a line in $\mathbb{R}^{n+k+1}$ and $\pi(l)$ is a line in $\mathbb{R}^{k+1}$. Now we specify a functional $\pi(l)\rightarrow \mathbb{R}$ by sending $(x_0,\ldots ,x_k)\mapsto y_1$. Likewise we obtain $n$ functionals on $\pi(l)$ via formulas $$(x_0,\ldots ,x_k)\mapsto y_1, \ \ \ \ldots \ \ \ (x_0,\ldots ,x_k)\mapsto y_n.$$
Finally we get continuous map $\mathbb{R}P^{n+k}-\mathbb{R}P^{n-1}\rightarrow nL^*$.
And I should check that this is the homeomorphism.
First of all to get a bijection we probably should get $(x_0,\ldots ,x_k)\mapsto y_i$ in some unique way, which I don't see.
Assuming that above is true and that multiplying all $x_i$'s and $y_i$'s by $\lambda$ gives rise to the same functional we obtain continuous bijection. But why the inverse is also continuous?