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There is no $m$. $k(n)$ was the upper limit for $i$ in the residue classes for which $\{x_i \pmod{p_i}\}$ was known if $p_i$ was the $i$-th prime number. See my counter example.
Any time you intersect $\bigcap_{i<n}\{x_i \pmod{p_i} \}$ you end up with a prime residue class and there are infinitely many primes with those $x_i$ for each $p_i$. You end up with the lowest possible value for $p_n$, and you can not determine it uniquely.
"You are claiming there are infinitely many prime number satisfying the congruence relation, which is true. But, $p_n$ is bounded by $k(n)$." And nowhere do you specify in the question (not the comment) that $p_n$ is bounded by $k(n)$.
