$$\begin{array}{l} a_{i} a_{n+1-i}=\left[a_{1}+(i-1) d\right]\left[a_{1}+(n-i) d\right] \\ =a_{1}^{2}+(n-1) a_{1} d+(i-1)(n-i) d^{2} \\ \geqslant a_{1}^{2}+(n-1) a_{1} d \\ =a_{1}\left[a_{1}+(n-1) d\right]=a_{1} a_{n}, \\ \quad\left(1+\sqrt[p]{a_{i}}\right)\left(1+\sqrt[p]{a_{n+1-i}}\right) \\ \quad=1+\left(\sqrt[p]{a_{i}}+\sqrt[p]{a_{n+1-i}}\right)+\sqrt[p]{a_{i} a_{n+1-i}} \\ \geqslant 1+2 \sqrt{\sqrt[p]{a_{i} a_{n+1-i}}}+\sqrt[p]{a_{i} a_{n+1-i}} \\ =\left(1+\sqrt{\sqrt[p]{a_{i} a_{n+1-i}}}\right)^{2} \\ \geqslant\left(1+\sqrt{\sqrt[p]{a_{1} a_{n}}}\right)^{2}=\left(1+\sqrt[2 p]{a_{1} a_{n}}\right)^{2} . \\ \quad \text { Hence } \prod_{i=1}^{n}\left[\left(1+\sqrt[p]{a_{i}}\right)\left(1+\sqrt[p]{a_{n+1-i}}\right)\right] \\ \geqslant\left(1+\sqrt[2 p]{a_{1} a_{n}}\right)^{2 n} . \end{array}$$

Therefore, $\prod_{i=1}^{n}\left(1+\sqrt[p]{a_{i}} \geqslant\left(1+\sqrt[2 p]{a_{1} a_{n}}\right)^{n}\right.$.
The equality holds when $a_{i}=a_{n+1-i}$, i.e., when the common difference of $\left\{a_{n}\right\}$ is zero.