jmc

Let $a_n$ denote the number of $n$-letter words ending in two constants (CC), $b_n$ denote the number of $n$-letter words ending in a constant followed by a vowel (CV), and let $c_n$ denote the number of $n$-letter words ending in a vowel followed by a constant (VC - the only other combination, two vowels, is impossible due to the problem statement). Then, note that: We can only form a word of length $n+1$ with CC at the end by appending a constant ($M,P$) to the end of a word of length $n$ that ends in a constant. Thus, we have the recursion $a_{n+1}=2(a_n+c_n)$, as there are two possible constants we can append. We can only form a word of length $n+1$ with a CV by appending $O$ to the end of a word of length $n$ that ends with CC. This is because we cannot append a vowel to VC, otherwise we'd have two vowels within $2$ characters of each other. Thus, $b_{n+1}=a_n$. We can only form a word of length $n+1$ with a VC by appending a constant to the end of a word of length $n$ that ends with CV. Thus, $c_{n+1}=2b_n$. Using those three recursive rules, and that $a_2=4,b_2=2,c_2=2$, we can make a table:$$\begin{array}{rrrr}& a_n& b_n& c_n\\ 2& 4& 2& 2\\ 3& 12& 4& 4\\ 4& 32& 12& 8\\ 5& 80& 32& 24\\ 6& 208& 80& 64\\ 7& 544& 208& 160\\ 8& 408& 544& 416\\ 9& 648& 408& 88\\ 10& 472& 648& 816\end{array}$$For simplicity, we used $\mathrm{mod}1000$. Thus, the answer is $a_{10}+b_{10}+c_{10}\equiv \boxed{936}(\mathrm{mod}1000)$.