I wanted to elaborate on coudy's original answer, but I also think something is wrong with this argument. In the absence of any other answers I will make the bounty available for an explanation of what goes wrong here...

If $\mathcal{P}$ is a partition of $M$, write $\mathcal{P}_m^\vee := \bigvee_{j=0}^m T^j \mathcal{P}$. (Because $T$ preserves the SRB measure $\mu$ we don't have to consider, e.g., $j<0$ terms.)

A Markov partition $\mathcal{R}$ is generating, so the supremum over partitions in the Kolmogorov-Sinai entropy is realized by it: the KS entropy is $h_\mu^{KS}(T) = -\lim_m m^{-1}\sum_{R \in \mathcal{R}_m^\vee} \mu(R) \log \mu(R)$. Meanwhile the topological entropy is $h(T) = \lim_m m^{-1} \log \# \mathcal{R}_m^\vee$.

So if $\mu$ is also the measure of maximal entropy (as is the case, e.g. when $T$ is a hyperbolic toral automorphism or the Poincaré map for a geodesic flow on a surface of negative curvature), then $h_\mu^{KS}(T) = h(T)$ and in the limit we have that $-\sum_{R \in \mathcal{R}_m^\vee} \mu(R) \log \mu(R) \sim \log \# \mathcal{R}_m^\vee$, so that $\mu$ is asymptotically uniform on $\mathcal{R}_m^\vee$.