Given,

$\overrightarrow{a}$ and $\overrightarrow{b}$ are two vectors such that $\overrightarrow{a}=2\hat{\mathrm{i}}+2\hat{\mathrm{j}}+p\hat{\mathrm{k}}, \left|\overrightarrow{b}\right|=7$

Now given $\overrightarrow{a}·\overrightarrow{b}=4$

So, applying the formula we get,

$\overrightarrow{a}·\overrightarrow{b}=4\Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta =4$

$\Rightarrow \sqrt{8+p^2}\times 7\cos \theta =4$

$\Rightarrow \cos \theta =\frac{4}{7\sqrt{8+p^2}}$

$\Rightarrow \sin \theta =\frac{\sqrt{{\left(7\sqrt{8+p^2}\right)}^2-4^2}}{7\sqrt{8+p^2}}=\frac{\sqrt{49p^2+376}}{7\sqrt{8+p^2}}$

Now also given $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=5\sqrt{17}$

$\Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin \theta =5\sqrt{17}$

$\Rightarrow \sqrt{8+p^2}\times 7\times \frac{\sqrt{49p^2+376}}{7\sqrt{8+p^2}}=5\sqrt{17}$

$\Rightarrow \sqrt{49p^2+376}=5\sqrt{17}$

$\Rightarrow 49p^2+376=25\times 17$

$\Rightarrow 49p^2=49$

$\Rightarrow p=\pm 1$