We are given with the linked list $L$ which contains $n$ elements. It is also mentioned that we have $\theta (n)$ space.
One thing that we can do is, traverse the whole list $L$, and save it to an array of size $n$, hence using that auxiliary space of $\theta (n)$. This will take $\theta (n)$ time.
Now traverse the list $P$ having $k$ elements, and for each element in $P$ print the element in that index in our array.
Say $P$ is $2,1,6$. Then we start traversing $P$, first is $2$, so we print $2nd$ element of our array, then $1st$ element, and then $6th$ element. And so on. This will take $\theta (k)$.
The whole procedure will take $\theta (n) + \theta (k)$, ie $ = \theta (n)$. Since, $k<<<n$.