We prove this generalization: If there are $n$ people each of $n$ different specialties, then the $n^2$ people can be seated around a round table such that, if $A$ and $B$ are two different people with the same specialty, then the people sitting to the immediate left of A and to the immediate left of $B$ are of different specialties. (Our problem is the case $n=5$.)

If $n=1$, then there is only one person, so the statement we're trying to prove is trivial. If $n=2$, then there are two people in each of two specialties. They can be seated so that both pairs of people of the same specialty are next to each other; this seating satisfies the requirements of the problem. We now continue by induction.

Suppose that we can seat $n$ people in each of $n$ specialties according to the conditions of the problem. Now consider the problem with $n+1$ people in each of $n+1$ specialties. To our existing $n$-specialty seating, we need to add $1$ additional person from each of the original $n$ specialties, plus $n+1$ people from the new specialty, which we'll call $Z$.

Let $S$ be one of the original $n$ specialties, and consider the left neighbors of the original $n$ people from $S$. Those neighbors must all have different specialties (among the original $n$ specialties), so among them there must be exactly one from each original specialty, including S itself. Thus there must be a pair of specialists in $S$ in adjacent seats. We seat the $(n+1)^{\text{th}}$ person from $S$ and one person from $Z$ (in either order) between the adjacent pair of people from $S$. We have thereby put a specialty-$Z$ person to the left of a specialty-$S$ person, and a specialty-$S$ person to the left of a specialty-$Z$ person; this is all we've done. So, it's still the case that no two people from the same specialty have people from the same specialty as their neighbors to the left. We repeat this operation for all $n$ choices of specialty $S$.

Finally, we need to add the $(n+1)^{\text{th}}$ specialty-$Z$ person. We can simply seat her next to any other specialty-$Z$ person. By construction, all the specialty-$Z$ people will have left neighbors of different specialties.

As an example, in our problem, we have $5$ people each of specialties M,B,C,P, and E. We can iterate the above process as follows, where the lists below are thought of as starting at some point on the table and going clockwise:

$\begin{align} M \\MMBB \\ MMCCMBBCB\\ MMPPMCCPCMBBPBCB\\MMEEMPPEPMCCECPCMBBEBPBCB \end{align}$

Note that in each list, no letter pair appears twice in the sequence.