Trigonometry Samurai!

$\begin{aligned} P & = \prod_{k=1}^{29} \left(\sqrt 3 + \tan k^\circ \right) \\ & = \prod_{k=1}^{29} \left(\sqrt 3 + \cot \left({\color{#3D99F6}90^\circ - k^\circ} \right) \right) & \small \color{#3D99F6} \text{For } k \in [1,29], 90-k \in [61, 89] \equiv 60+k \in [61, 89] \\ & = \prod_{k=1}^{29} \left(\sqrt 3 + \cot \left({\color{#3D99F6}60^\circ + k^\circ} \right) \right) \\ & = \prod_{k=1}^{29} \left(\sqrt 3 + \frac {1-\tan 60^\circ \tan k^\circ}{\tan 60^\circ + \tan k^\circ} \right) \\ & = \prod_{k=1}^{29} \left(\sqrt 3 + \frac {1-\sqrt 3 \tan k^\circ}{\sqrt 3 + \tan k^\circ} \right) \\ & = \prod_{k=1}^{29} \frac 4{\sqrt 3 + \tan k^\circ} \end{aligned}$

Now, we have:

$\begin{aligned} P^2 & = \prod_{k=1}^{29} \left(\sqrt 3 + \tan k^\circ \right) \prod_{k=1}^{29} \frac 4{\sqrt 3 + \tan k^\circ} = \prod_{k=1}^{29} 4 = 4^{29} \\ \implies P & = 2^{29} \end{aligned}$

Therefore, $a+b = 2+29 = \boxed{31}$ .