I'm looking at the proof of Theorem 3.4.4 in "Frank Stenger Numerical Methods based on Sinc and Analytic Functions". The book seems to be available on google books, but I will try to make the specific issue self-contained.
Let $t,d,h \in \mathbb{R}$ and $z\in \mathbb{C}$, write $z=x + iy$. Assume $-d<y<d$. Then I need the bound $$ \left \lvert \frac{ e^{-i \pi (t-id)/h} - \cos(\pi z /h)}{ (t-id-z) \sin( \pi (t-id)/h)}-\frac{ e^{-i \pi (t+id)/h} - \cos(\pi z /h)}{ (t+id-z) \sin( \pi (t+id)/h)} \right \lvert \leq \frac{ e^{-\pi d/h} + \cosh(\pi y/h)}{ d \sinh(\pi d / h)} $$
It is clear that $\lvert e^{-i \pi (t-id)/h} \lvert \leq e^{- \pi d /h}$ and that $\lvert \cos(\pi z /h)) \lvert \leq \cosh(\pi y/h)$, so the triangle inequality gets us some way. I can't seem to get anywhere with the denominator though.
First $|t\pm id-z|\ge |d|$. Also for real $a$ and $b$ $$ |\sin(a+ib)|=|\sin a \cosh b+i \cos a \sinh b|=|\sin a\, \text{coth}\, b+i \cos a| |\sinh b |\ge |\sinh b| $$ since $|\text{coth}\, b|\ge 1$. This leads to the rhs $$ 2\frac{ e^{-\pi d/h} + \cosh(\pi y/h)}{ |d \sinh(\pi d / h)|}. $$