\(f(x) = - (p^2 - 6p + 8) \cos 4x + 2(2 - p)x + 7\)
\(f'(x) = +4 \left( p^2 - 6p + 8 \right) \sin 4x + (4 - 2p) \neq 0\)
\(\sin 4x \neq \frac{2p - 4}{4(p-4)(p-2)}\)
\(\sin 4x \neq \frac{2(p-2)}{4(p-4)(p-2)}\)
\(p \neq 2\)
\(\sin 4x = \frac{1}{2(p-4)}\)
\(\implies \left| \frac{1}{2(p-4)} \right| > 1\)
On solving, we get:
\(\therefore p \in \left( \frac{7}{2}, \frac{9}{2} \right)\)
Hence \(a = \frac{7}{2}\), \(b = \frac{9}{2}\).
\(\therefore 16ab = 252\)