We have (i) $k=0.5\, p_N$ works for $N$ large enough, (ii) $k=p_N$ works for all $N$ and (iii) $k\le 0.3 \, p_N$ will fail if
$p_{N+2}=p_{N+1}+2$ and $N$ is large.

With $q=p_{N+1}$, we have
$$
\frac{1}{1-q^{-2k}} < \frac{1}{1-x^{-2k}} = \frac{1}{1-q^{-2k}} \prod_{p>q} \frac{1}{1-p^{-2k}} .
$$
It follows that 
$$
q^{-2k} < x^{-2k} < q^{-2k} + \sum_{j\ge 2} (q+j)^{-2k} < q^{-2k} +\frac{1}{(q+1)^{2k-1}(2k-1)}.
$$
Taking logarithms, and using $\log(1+y)\le y$, we get
$$
-2k \log q < -2k \log x < -2k\log q + \frac{q+1}{\exp\{(1-o(1)) 2k/q\} (2k-1)}.
$$
Dividing by $-2k$ and exponentiating, we have
$$
q > x > q - \frac{q(q+1)}{\exp\{(1-o(1))2k/q\} 2k (2k-1)}.
$$
We want the last expression to be less than $1/2$. Since $q/p_N \to 1 $ as $N\to \infty$, we need $k\ge (1+o(1)) c \, p_N$,
where $c=0.45...$ is the solution to $e^{2c}4 c^2=2$.
So $k=0.5 \, p_N$ works for $N$ large enough, while $k=p_N$ likely works for all $N$. 
Similar calculations show that when $p_{N+2}=p_{N+1}+2$ and $N$ is large, it is necessary that $k>0.3 \, p_N$.