I'm not sure about your definition, so let's take branch number in terms of the byte-wise differential branch number, i.e. the branch number of a function $F(x)$ is $$\mathcal{B}_{F(x)} = \min_{a,b \neq a}\{ w(a \oplus b) + w(F(a) \oplus F(b))\}$$ where $w(x)$ is the number of non-zero bytes in $x$.
In this case, the branch number of the Twofish round function can be no greater than 5.
To see why, consider the situation where the PHT only has a branch number of 2 - i.e. if you start the PHT with only the most significant bit of word $A$ being differentially active (defining the PHT as $PHT(A,B) = 2A+B,A+B$). In that case, after the PHT only a single byte is active (the byte containing the most significant bit of $A+B$).
Now go back to the situation just before the PHT and think about performing the round function in reverse. There is a single 32-bit word that is active, and of that word only a single byte is active. Applying the MDS layer in reverse, this translates into 4 bytes being differentially active at the start of the round function. So the Twofish round function has a branch number of at most 4+1 bytes.
But ... is the branch number less than 5? That I don't know, though it would seem difficult for it to be so given that the branch number of each MDS is 5. There would have to be a scenario where few active bytes before the MDS translate into many active bytes prior to the PHT which then translates into few active bytes after the PHT. Ruling out such a scenario would prove that the branch number was in fact 5.
