Here the given expression is $\left(\frac{p}{2}+2\right)^8$.
$\therefore n=8$ [even]
$\therefore$ the Binomial expansion of the given expression has only one middle term i.e., $\left(\frac{8}{2}+1\right)$ th $=5$ th term
$$
\therefore T_5=T_{4+1}={ }^8 C_4\left(\frac{p}{2}\right)^{8-4} \cdot 2^4
$$


$$
\begin{aligned}
&\Rightarrow 1120={ }^8 C_4 p^4 \cdot 2^{-4} 2^4 \\
&\Rightarrow 1120=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 4 \times 3 \times 2 \times 1} p^4 \\
&\Rightarrow 1120=7 \times 2 \times 5 \times p^4 \\
&\Rightarrow p^4=\frac{1120}{70}=16 \Rightarrow p^4=2^4 \\
&\Rightarrow p^2=4 \Rightarrow p=\pm 2
\end{aligned}
$$