a.
Note that $n-k = 2p_{k} - k < 2p_{k} - p_{k} = p_{k}$, so $p_{k} \nmid (n-k)!$, so $2p_{k} \nmid (n-k)!$.

b.
Note that $n > 2p_{k} \geq 3k/2$ implies $k < 2n/3$, so $n-k > n/3$. So if we can find integers $a, b \geq 3$ such that $n = ab$ and $a \neq b$, then both $a$ and $b$ will appear separately in the product $(n-k)! = 1 \times 2 \times \cdots \times (n-k)$, which means $n \mid (n-k)!$. Observe that $k \geq 14$ implies $p_{k} \geq 13$, so that $n > 2p_{k} \geq 26$.

If $n = 2^{\alpha}$ for some integer $\alpha \geq 5$, then take $a = 2^{2}$, $b = 2^{\alpha-2}$.

Otherwise, since $n \geq 26 > 16$, we can take $a$ to be an odd prime factor of $n$ and $b = n/a$, unless $b < 3$ or $b = a$.

Case (i): $b < 3$. Since $n$ is composite, this means $b = 2$, so that $2a = n > 2p_{k}$. As $a$ is a prime number and $p_{k}$ is the largest prime number which is strictly less than $k$, it follows that $a \geq k$. From $n-k = 2a - k \geq 2a - a = a > 2$ we see that $n = 2a$ divides into $(n-k)!$.

Case (ii): $b = a$. Then $n = a^{2}$ and $a > 6$ since $n \geq 26$. Thus $n-k > n/3 = a^{2}/3 > 2a$, so that both $a$ and $2a$ appear among $\{1, 2, \ldots, n-k\}$. Hence $n = a^{2}$ divides into $(n-k)!$.