The case $k=2.5$ is trivial. There is counterexample! Let $k=3$. Without loss of generality let $P_1 \equiv (-3,0)$ and $P_n \equiv (3,0)$. Let $M$ be $S$ under translation by the vector $-\frac{1}{3} . \vec{P_1P_n}$. Let $N$ be $S$ under translation by the vector $\frac{1}{3} . \vec{P_1P_n}$. Suppose that we can't find $A$ and $B$ such that $k=3$ and $AB||P_1P_n$. This means that $M\cap S=\emptyset$ and $N\cap S=\emptyset$ Now we have three broken lines $M,S$ and $N$. Let's continue $S$ in both directions so that we divide the plane into two parts $\mu$ and $\nu$, and M is in the part $\mu$ and N is in the part $\nu$. Lets look at the $Ox$ axis. Each point of it is either in $\mu$ or $\nu$ or in the border $S$. The point $-3$ is on the border, so $-1$ is in $\nu$. $3$ is on the border, so $1$ is in $\mu$. So there exist $-1\leq B_1 \leq 1$ so that $B_1$ is on the border. So $-3\leq B_1 -2\leq -1$ and $M_1=B_1 -2$ is in $\mu$. So there exist $M_1\leq B_2 \leq N$ with $B_2$ on the border. Then $N_2= B_2 +2$ is between $B_1$ and $1$. There exist $B_3$ between $N_2$ and $1$ on the border and we build infinite sequence $B_{2k+1}$ of points on the border and between every two there is point not on the border. That means that the broken line S intersects the $Ox$ axis infinite number of times- \textbf{CONTRADICTION}

imortal wrote: The case $ k = 2.5$ is trivial. There is counterexample! How?

If $k\ge 0$, this is always possible iff $k$ is a nonnegative integer. If $k\le1$, this is all obvious; from now on, suppose that $k>1$. First suppose $k\notin\mathbb{Z}$, and let $m=\lceil{k}\rceil\ge2$ so that $m-1<k<m$. For $1\le i\le m-1$, we take \begin{align*} &P_{8i-7}=(m(i-1),0), P_{8i-6}=(m(i-1),m-i), P_{8i-5}=((m-1)i,m-i), P_{8i-4}=((m-1)i,0), \\ &P_{8i-3}=((m-1)i,0), P_{8i-2}=((m-1)i,-i), P_{8i-1}=(mi,-i), P_{8i}=(mi,0), \end{align*}where the fact that $m(i-1)<(m-1)i<mi$ ensures that $S$ is a simple curve (although the problem does not stipulate this, it's convenient for the construction). It's easy to see that $P_1=(0,0)$, $P_n=(m(m-1),0)$, and the resulting figure $S$ is symmetric about $(m(m-1)/2,0)$. Hence if we go by contradiction (i.e. assume we can find $A,B$ such that $AB\in(m-1,m)$), we can WLOG work with $A,B$ on or above the $x$-axis. Note that the region of $S$ above the $x$-axis is a sequence of $m-1$ disjoint squares (bottom sides missing) with decreasing height (from left to right), so we can assume that $A$ lies on the $i^{th}$ square and $B$ lies on the $j^{th}$, where $1\le i<j\le k-1$. Since $A,B$ have the same $y$-coordinate, the $x$-coordinate of $A$ is either $m(i-1)$ or $(m-1)i$, and the $x$-coordinate of $B$ lies in the closed interval $[m(j-1),(m-1)j]$. If $A_x=m(i-1)$, then $AB\ge m(j-1)-m(i-1)\ge m$, contradiction, so \[A_x=(m-1)i\implies AB\in[m(j-1)-(m-1)i,(m-1)(j-i)].\]But if $j=i+1$, then $AB\le m-1$, and if $j\ge i+2$, then $m(j-1)-(m-1)i\ge m(i+1)-(m-1)i=m+i\ge m$, contradiction. It remains to show that when $k\ge2$ is an integer, there exist $A,B\in S$ such that $P_1P_n=kAB$. Suppose otherwise; WLOG set $P_1=(0,0)$ and $P_n=(k,0)$. For any curve $C$ (possibly a point) and real $r$, let $C^{(r)}=C+(r,0)$. By assumption, $C^{(r)}$ and $C^{(r+1)}$ are disjoint for all $r$. The key idea is to define "left/right" properly: take indices $M$ and $m$ such that the $y$-coordinates of $P_M$ and $P_m$ are maximal and minimal, respectively, and $|M-m|$ is as small as possible; WLOG $m<M$, and $J$ is the region bounded by the two horizontal lines through $P_M,P_m$ (inclusive); note that $y=0$ lies in $J$. Let $T$ denote the broken line $P_mP_{m+1}\ldots P_M$, the exterior $E$ be the set of points $X\ne P_M\in J$ such that there exists a continuous path from $P_M$ to $X$ intersecting $T$ only at the beginning (at $P_M$), and the interior $I=J\setminus E$. Define the left $L$ of $T$ as the set of points $X\in J$ such that there exists a continuous path from $(-\infty,0)$ to $X$ that stays in $J$ and never meets $T$; define the right $R$ analogously (note that the points in $I$ are in neither $L$ nor $R$). Clearly $E=L\cup R$. If $S^{(1)}$ intersects both $I$ and $E$, then $S^{(1)}$ clearly hits $T$, contradiction, and if $S^{(1)}$ lies entirely in $I$, then the uppermost $y$-coordinate in $S^{(1)}$ is strictly less than that of $I$ (since $S^{(1)}$ does not intersect $T$), contradicting the fact that $P_M^{(1)}\in S^{(1)}$. Hence $S^{(1)}$ lies entirely in either $L$ or $R$ (since $S^{(1)}$ never leaves $J$ and never hits $T$). But it's not hard to show that $P_M^{(1)}\in R$, so $S^{(1)}$ lies entirely in $R$. In particular, $I^{(1)}\subset R$, so $R^{(1)}\subset R$ by the continuity of $I^{(1)}$ and $R$ (in other words, a path staying in $J$ from $(\infty,0)$ to some $X\in R^{(1)}$ cannot meet $I$ without first hitting $I^{(1)}$, since $I$ and $I^{(1)}$ are disjoint). Hence $S^{(2)}\subset R^{(1)}\subset R$; by a simple induction, $S^{(i)}$ lies entirely in $R$ for $i\ge1$, and by symmetry, $S^{(-i)}$ lies entirely in $L$ for $i\ge1$. Thus $P_1^{(k-1)}\in R$ (recall that $k\ge2$) and $P_n^{(-1)}\in L$, contradicting the fact that $P_1^{(k-1)}=P_n^{(-1)}$ and $L\cap R=\emptyset$.