10. (15 points) As shown in Figure 3, three circles intersect to form seven regions. Fill in the numbers $1, 2, 2, 4, 4, 8, 8$ into these seven regions so that the product of the four numbers inside each circle is equal (denoted as $P$). The filling method shown in Figure 4 satisfies the condition, with $P=64$ at this time. For all filling methods that meet the above requirements, find the maximum and minimum values of $P$.


10. Let $a \sim g$ represent a permutation of $1,2,2,4,4,8,8$ that satisfies the conditions of the filling method in Figure 7. Then $P=a b c e = a b d f = a c d g$. Note that,
$$
\text { abcdefg }=1 \times 2 \times 2 \times 4 \times 4 \times 8 \times 8=2^{12} \text {. }
$$

Thus, $P^{3}=a^{3} b^{2} c^{2} d^{2} efg =2^{12} a^{2} b c d$. Since $P^{3}$ is a cube number, $a^{2} b c d$ must also be a cube number. And $a^{2} b c d$ is a positive integer power of 2, with
$$
\begin{array}{l}
2^{4}=1^{2} \times 2 \times 2 \times 4 \leqslant a^{2} b c d \\
\leqslant 8^{2} \times 8 \times 4 \times 4=2^{13},
\end{array}
$$

Therefore, $2^{6} \leqslant a^{2} b c d \leqslant 2^{12}$. Figures 4 and 8 show the filling methods that make $a^{2} b c d=2^{6}$ and $2^{12}$, respectively. Thus, $P_{\max }=\sqrt[3]{2^{12} \times 2^{12}}=256$,
$$
P_{\min }=\sqrt[3]{2^{12} \times 2^{6}}=64 .
$$