Following a more elementary route we will first break the integral in two pieces:
$$B(\frac{N+1}{N+2};N+1,p+1)=B(N+1,p+1)-I\\I=\int^{1}_{\frac{N+1}{N+2}}x^{N}(1-x)^pdx$$
Perform the following change of variables $x=1-\frac{t}{N+2}$:
$$I=\frac{1}{(N+2)^{p+1}}\int_{0}^1t^p(1-\frac{t}{N+2})^{N}dt$$ However it is possible to estimate that
$$(1-\frac{t}{N+2})^{N}=e^{-t}(1+\frac{4t-t^2}{2N}+\mathcal{O}(N^{-2}))$$
and performing the integrals we get:
$$I\sim\frac{1}{N^{p+1}}\Big[\gamma(p+1,1)+\frac{2}{N}(\gamma(p+2,1)-p-1-\frac{1}{4}\gamma(p+3,1))\Big]$$
While on the other hand we may write
$$B(N+1,p+1)\sim \frac{\Gamma(p+1)}{N^{p+1}}\Big[1-\frac{(p+1)^2+3(p+1)}{2N}\Big]$$
So all in all as a leading approximation one gets:
$$B(\frac{N+1}{N+2};N+1,p+1)\sim\frac{\Gamma(p+1)-\gamma(p+1,1)}{N^{p+1}}=\frac{\Gamma(p+1,1)}{N^{p+1}}+\mathcal{O}(N^{-p-2})$$
where $\Gamma(z,x)=\int_x^{\infty}t^{z-1}e^{-t}dt$ is the upper incomplete Gamma function. From the above one can calculate also the subleading term. More subleading terms can also be readily calculated but probably not in closed form.
By this known result
$$
\mathrm{e}^{ - x\pi /2} \left| {\Gamma \!\left( {h \pm \mathrm{i}\frac{x}{2} } \right)} \right|^2 \sim 2\pi \left( {\frac{x}{2}} \right)^{2h - 1} \mathrm{e}^{ - x\pi}
$$
as $x\to +\infty$ with any fixed $h>0$. As $x\to 0$, we have
$$
\mathrm{e}^{ - x\pi /2} \left| {\Gamma \!\left( { h\pm \mathrm{i}\frac{x}{2} } \right)} \right|^2 = \Gamma ^2 (h) - \frac{\pi }{2}\Gamma ^2 (h)x + \mathcal{O}_h(x^2 ).
$$
The absolute value of the gamma function is an even function, thus the more complicated behaviour is hidden in the $\mathcal{O}$-term. Only the exponential function contributes to the linear term.
Best Answer
We can write $$ f(b) = 2b^{1 - 1/m} \Gamma \left( {1 - \frac{1}{m}} \right)\operatorname{Re} \left( {\mathrm{e}^{\frac{{\mathrm{i}\pi }}{{2m}}} \frac{{\Gamma \left( { - \mathrm{i}b + \frac{1}{{2m}}} \right)}}{{\Gamma \left( { - \mathrm{i}b + 1 - \frac{1}{{2m}}} \right)}}} \right). $$ It is shown in this paper that \begin{align*} \log \Gamma ( - \mathrm{i}y + a) \sim \left( { - \mathrm{i}y + a - \frac{1}{2}} \right)\log (y\mathrm{e}^{ - \frac{\pi }{2}\mathrm{i}} ) + \mathrm{i}y + \log \sqrt {2\pi } & + \sum\limits_{k = 1}^\infty {\frac{{\mathrm{e}^{ - 2\pi \mathrm{i}ka} }}{{2k}}\mathrm{e}^{ - 2\pi ky} } \\ & - \sum\limits_{n = 2}^\infty {\frac{{B_n (a)}}{{n(n - 1)(\mathrm{i}y)^{n - 1} }}} \end{align*} as $y\to +\infty$ with any fixed $0\leq a\leq 1$. Here $B_n(a)$ are the Bernoulli polynomials. The importance of this expansion is that it takes into account the exponentially small contributions on the Stokes lines for $\Gamma(z+a)$, which is crucial in getting the right asymptotics for $f(b)$. Consequently, \begin{align*} \log \frac{{\Gamma \left( { - \mathrm{i}b + \frac{1}{{2m}}} \right)}}{{\Gamma \left( { - \mathrm{i}b + 1 - \frac{1}{{2m}}} \right)}} \sim\; & \left( {\frac{1}{m} - 1} \right)\log (b\mathrm{e}^{ - \frac{\pi }{2}\mathrm{i}} ) + \sum\limits_{k = 1}^\infty {\frac{{\mathrm{e}^{ - \pi \mathrm{i}k/m} - \mathrm{e}^{\pi \mathrm{i}k/m} }}{{2k}}\mathrm{e}^{ - 2\pi kb} } \\ & + \sum\limits_{n = 2}^\infty {\frac{{B_n \left( {1 - \frac{1}{{2m}}} \right) - B_n \left( {\frac{1}{{2m}}} \right)}}{{n(n - 1)(\mathrm{i}b)^{n - 1} }}} \\ = \; & \left( {\frac{1}{m} - 1} \right)\log b - \frac{\pi }{2}\left( {\frac{1}{m} - 1} \right)\mathrm{i} - \mathrm{i}\sum\limits_{k = 1}^\infty {\frac{{\sin (\pi k/m)}}{k}\mathrm{e}^{ - 2\pi kb} } \\ & - \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n B_{2n + 1} \left( {\frac{1}{{2m}}} \right)}}{{n(2n + 1)b^{2n} }}} \\ \\ = \; & \left( {\frac{1}{m} - 1} \right)\log b - \frac{\pi }{2}\left( {\frac{1}{m} - 1} \right)\mathrm{i} - \mathrm{i}\sum\limits_{k = 1}^\infty {\frac{{\sin (\pi k/m)}}{k}\mathrm{e}^{ - 2\pi kb} }\\ & + \mathcal{O}\!\left( {\frac{1}{{b^2 }}} \right), \end{align*} as $b\to +\infty$, with the big-$\mathcal{O}$ function being real valued. Using this result in the above form for $f(b)$ and simplifying, we find \begin{align*} f(b) & = 2\Gamma \left( {1 - \frac{1}{m}} \right)\sin \left( {\sum\limits_{k = 1}^\infty {\frac{{\sin (\pi k/m)}}{k}\mathrm{e}^{ - 2\pi kb} } } \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{{b^2 }}} \right)} \right) \\ & = 2\Gamma \left( {1 - \frac{1}{m}} \right)\sin \left( {\frac{\pi }{m}} \right)\mathrm{e}^{ - 2\pi b} \left( {1 + \mathcal{O}\!\left( {\frac{1}{{b^2 }}} \right)} \right) \\ & = \frac{{2\pi }}{{\Gamma \left( {\frac{1}{m}} \right)}}\mathrm{e}^{ - 2\pi b} \left( {1 + \mathcal{O}\!\left( {\frac{1}{{b^2 }}} \right)} \right) \end{align*} as $b\to+\infty$.