It is impossible. The proof goes here.
Denote $d$ as the order of $p_0$, so $d$ divides the order of $\mathbb{Z}^\times_N$ that is $4p'q'$. Let, $e=p_1\times p_2\times\cdots\times p_m$. Since, $p_i$s are odd, $e$ is odd.
If $p_0^e=1$ then $d$ divides $e$. So, $d$ divides $\mathsf{gcd}(e,4p'q')=1$ which implies $d=1$. In $\mathbb{Z}^\times_N$, the only element with order $d=1$ is the identity $1$ itself. But, $p_0>1$, so, $p_0^e\ne 1$.
Similarly, if $p_0^e=-1$ then $p_0^{2e}=1$, so $d$ divides $2e$. So, $d$ divides $\mathsf{gcd}(2e,4p'q')=2$ which implies $d=2$. It means either $p_0=-1 \pmod{p}$ or $p_0=-1 \pmod{q}$ or both i.e., $p_0=-1 \pmod{N}$. Hence, $p_0\ge p-1=2p'$ or $p_0\ge q-1=2q'$. But, $p_0< p'$ and $p_0<q'$. Hence, $p_0^e\ne-1$.