Sky is the limit!

$\begin{aligned} P & = \prod_{\alpha = 1}^{99} \lim_{n \to \infty} \prod_{k=1}^{\alpha n} \sqrt [n] {1+\frac nk} \\ & = \prod_{\alpha = 1}^{99} \lim_{n \to \infty} \exp \left(\sum_{k=1}^{\alpha n} \ln \left(\sqrt [n] {1+\frac nk}\right) \right) \\ & = \prod_{\alpha = 1}^{99} \exp \left({\color{#3D99F6}\lim_{n \to \infty} \sum_{k=1}^{\alpha n} \frac 1n \ln \left(1+\frac 1{\frac kn} \right)} \right) \quad \quad \small \color{#3D99F6} \text{ Riemann's sum: } \lim_{n \to \infty} \sum_{k=a}^b \frac 1n f\left(\frac kn\right) = \int_{\lim_{n \to \infty} a/n}^{\lim_{n \to \infty} b/n} f(x) \ dx \\ & = \prod_{\alpha = 1}^{99} \exp \left({\color{#3D99F6} \int_0^\alpha \ln \left(1+\frac 1x \right) dx} \right) \\ & = \prod_{\alpha = 1}^{99} \exp \left(\int_0^\alpha \ln \left(\frac {x+1}x \right) dx \right) \\ & = \prod_{\alpha = 1}^{99} \exp \left(\int_0^\alpha \ln (x+1) - \ln x \ dx \right) \\ & = \prod_{\alpha = 1}^{99} \exp \left((x+1)(\ln (x+1) -1) - x (\ln x - 1) \bigg|_0^\alpha \right) \\ & = \prod_{\alpha = 1}^{99} \exp \left((x+1)\ln (x+1) -1 - x \ln x) \bigg|_0^\alpha \right) \\ & = \prod_{\alpha = 1}^{99} \exp \left( \ln \left(\frac {(\alpha+1)^{\alpha+1}}{\alpha^\alpha} \right) \right) \\ & = \prod_{\alpha = 1}^{99} \frac {(\alpha+1)^{\alpha+1}}{\alpha^\alpha} = \frac { \prod_{\alpha = 1}^{99} (\alpha+1)^{\alpha+1}}{ \prod_{\alpha = 1}^{99} \alpha^\alpha} \\ & = \frac { \prod_{\alpha = 2}^{100} \alpha^\alpha}{ \prod_{\alpha = 1}^{99} \alpha^\alpha} = \frac {100^{100}}{1^1} = 100^{100} \end{aligned}$

$\implies$ the minimum $a+b = 100+100 = \boxed{200}$