The homothetical stuff is useless. Step one. The expression shouts $1+ \binom{n}{2}\frac{1}{kn}$.. so box principle it is. Step two. Given polygons $P_i$ and $P_j$ at least one polygon (Wlog $P_i$) has one of its vertexes contained in the other (bound+ interior).In this case we say $P_i$ gives $P_j$ Step three We have $\binom{n}{2}$ pairs $P_i,P_j$ so there is one which gives $\binom{n}{2}\frac{1}{n}$ vertexes. Since he has k vertexes there is a vertex which is given to at least $\binom{n}{2}\frac{1}{kn}$ polygons. Thus that vertex $V$ is contained in $1+ \binom{n}{2}\frac{1}{kn}$ (plus one because of the polygon who gives).

Lemma. If two such polygon $P_i$ and $P_j$ intersects then one vertex of them belongs to the other. Proof. Let their homothetic center be $O$. Suppose the homothety sending $P_i$ to $P_j$ has ratio $k>1$. Let $P^t_i$ be the polygon defined as the image of $P_i$ under a homothety at $O$ with ratio $t$. Since the polygon is convex we see that the values of $t$ such that $P_i$ and $P_j$ intersect is a closed interval $[1,m]$. Notice that at $t=m$ the intersection of $P_i^{t}$ and $P_i$ must be a single point, hence a vertex of either $P_i^t$ or $P_i$. Therefore, this vertex must lie in both $P_i$ and $P_j$. $\blacksquare$ Now we count the number $N$ of pairs $(V_i,P_j)$ where $V_i$ is a vertex of a polygon $P_i$ which belongs to another polygon $P_j$. From the lemma, $(i,j)$ and $(j,i)$ contributes at least $1$ pair combined, hence $N\geq \binom{n}{2}$. Meanwhile, there are at least $nk$ vertices. Therefore, one of them must belong to $$\frac{\binom{n}{2}}{nk}=\frac{n-1}{2k}$$vertices not including the one that it belongs to as a vertex. Therefore, it belongs to at least $$1+\frac{n-1}{2k}$$vertices.