How do I find running time for the following divide and conquer problem?
You cannot use the Master Theorem to solve this recurrence, since $n!$ doesn't grow like $n^c$ for any $c$. Instead, Stirling's approximation shows that $n! \sim \sqrt{2\pi n} (n/e)^n$. Rather, expand the recurrence to get
$$
T(n) = n! + 16 (n/4)! + 16^2 (n/16)! + \cdots.
$$
Stirling's approximation shows that the latter terms are tiny compared to the first one. Therefore
$$
T(n) \sim n!.
$$
This is just shorthand for
$$
\lim_{n \to \infty} \frac{T(n)}{n!} = 1.
$$

If you want to prove this formally, one way is as follows. Start with the observation that $n/16 \geq 4$,
$$
16 (n/4)! \leq (n/4+1)(n/4)! = (n/4+1)! \leq \frac{1}{n} n!.
$$
The first inequality uses $n \geq 60$, and the second $n \geq 4$. Continuing in this vain, for $n/16^2 \geq 4$,
$$
16^2 (n/16)! \leq \frac{1}{n/4} (n/4)! \leq \frac{1}{n/4} \frac{1}{n} n! \leq \frac{1}{n} n!.
$$
In the same way we show that $16^k (n/4^k) \leq n!/n$ for $n/4^k \geq 4$. Therefore
$$
T(n) = n! + 16(n/4)! + 16^2(n/4^2)! + \cdots \leq n! + n!/n + n!/n + \cdots + 16^r T(4),
$$
where $n/4^r = 4$, and so $r = \log_4 n - 1$, which implies $16^r = n^2/16$. Since there are $r-1 = \log_4 n - 2$ summands of the form $n!/n$, we certainly have
$$
T(n) \leq n! + \frac{\log_4 n}{n} n! + \frac{T(4)}{16} n^2.
$$
This shows that $T(n) = n! + f(n)$, where
$$
f(n) = \frac{\log_4 n}{n} n! + \frac{T(4)}{16} n^2
$$
is a function satisfying
$$
\lim_{n \to \infty} \frac{f(n)}{n!} = 0.
$$
On the other hand, clearly $T(n) \geq n!$. This implies what we claimed above, $T(n) \sim n!$.