In the expression below, each letter corresponds to a digit, with different letters corresponding to different digits. Determine these digits.

$$
6 \times A O B M E P=7 \times M E P A O B
$$
Observe that

$$
A O B M E P=A O B \times 1000+M E P \text { and } M E P A O B=M E P \times 1000+A O B
$$

Let $A O B=m$ and $M E P=n$. From the problem, we have

$$
6 \times A O B M E P=7 \times M E P A O B
$$

which gives $6 \cdot(1000 m+n)=7 \cdot(1000 n+m)$, or $6000 m-7 m=7000 n-6 n$ or, equivalently, $5993 m=6994 n$. Dividing both sides by 13, we conclude that $461 m=538 n$. The prime factorization of 538 is $538=2 \times 269$ and 461 is prime. Therefore, 538 and 461 are coprime. Thus, 461 divides $n$ and 538 divides $m$. Since $A O B$ and $M E P$ are three-digit numbers, we can only have the solutions $n=461$, or $n=822$, and $m=538$. It is easy to verify that $6 \times 538461=3230766=7 \times 461538$ and that $6 \times 538822=3232932 \neq 5757766=7 \times 822538$. Therefore, $n=822$ does not work, and $A O B=538$ and $M E P=461$ is the only solution. Thus, the digits are $A=5, O=3, B=8, M=4, E=6$ and $P=1$.