First consider to functions , say $a_n,b_n$ and $a_n=\min(v_1(n),v_2(n),......,v_p(n)$ similarly call $b_n$ as max of those. Now if $a_n=b_n$ then just taking $N=n,a_n=k$ we're done. If not then :- Lemma : When $f(n)<g(n)$ , there exist $x>0$ such that $a_n\leq a_{n+x}\leq b_{n+x}<b_n$. Proof : First of all observe the sequence say $A_{n+m}=(V_m|[s(m)=v_m(n+m)]=a_n$ is decreasing for $m\geq 1$. So clearly we've $s(x+1)$ is subset of $s(x)$. Suppose $V_l$ be a linked vertex to $V_i$ in the same face with $V_s$. Clearly there exists a third vertex $V_s$ such that $v_s(n+x-1)=v_i(n+x)=a_n$ since $V_s$ was chosen randomly. Now in the case $A_{n+x}\ne \phi$ the above inclusion is strict. But according to given condition all vertices are contained in $A_{n+x}$. Clearly absurd. So lemma proved. Similarly we can also show for show for $b_n$ Back to main proof : From this surely there must have $y$ such that $a_m=b_m$ for some $m\geq n+x$. Now taking $a_m=k$. Done.