Let $f(x) = \sum_{k=2}^{p} \lfloor \log_k x \rfloor$ and let $M = \{f(x) \mid x \in [a, a+n]\}$. It is easy to show that if $k \ge 2$ is a positive integer, then $\lfloor \log_k \lfloor x \rfloor \rfloor = \lfloor \log_k x \rfloor$. This implies that $f(x) = f(\lfloor x \rfloor)$, for all $x \in [1, \infty)$, and hence $M = \{f(x) \mid x \in S\}$, where $S = \{\lfloor a \rfloor, \lfloor a \rfloor + 1, \dots, \lfloor a \rfloor + n\}$ has $n+1$ elements. On the other hand, for $s \in S$, $s < \lfloor a \rfloor + n \le p$, we have $s+1 \in \{2, 3, \dots, p\}$, and
$$f(s+1)-f(s) = \sum_{k=2}^{p} (\lfloor \log_k(s+1) \rfloor - \lfloor \log_k s \rfloor) \ge \lfloor \log_{s+1}(s+1) \rfloor - \lfloor \log_{s+1} s \rfloor = 1,$$
therefore $f(s+1) > f(s)$, and this proves that $M$ has exactly $n+1$ elements.