Let the set of all values of $p$, for which $f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right)+2(2-p) x+7$ does not have any critical point, be the interval $(a, b)$. Then $16 a b$ is equal to _________.

<p>$$\begin{aligned} & f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right) +2(2-p) x+7 \\ & f(x)=-\cos 4 x\left(p^2-6 p+8\right)+2(2-p) x+7 \\ & f^{\prime}(x)=4 \sin 4 x\left(p^2-6 p+8\right)+2(2-p) \neq 0 \\ & 2(2-p)+\left[-4\left(p^2-6 p+8\right), 4\left(p^2-6 p+8\right)\right] \\ & \Rightarrow\left[-4 p^2+24 p-32,4 p^2-24 p+32\right]+(4-2 p) \\ & {\left[-4 p^2+22 p-28,4 p^2-26 p+36\right]} \\ & {[(p-2)(-4 p+14),(p-2)(4 p-18)]} \\ & \Rightarrow(p-2)[(-4 p+14), 4 p-18] \Rightarrow p \in\left(\frac{7}{2}, \frac{9}{2}\right) \\ & \Rightarrow a=\frac{7}{2}, b=\frac{9}{2} \\ & \Rightarrow 16 a b=4 \times 63=252 \end{aligned}$$</p>