We have the Lagrangian: $$L(\theta, \phi, \dot{\theta}, \dot{\phi})=\frac{mb^2}{2}(\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2)+mbgcos(\theta)$$ And I have found that $L(\theta, \phi, \dot{\theta}, \dot{\phi})$ is strictly convex for $\theta \in \mathbb R \backslash \{0\}$ and for $\phi \in \mathbb R$) and I have found that $p_{\phi}=mb^2sin^2(\theta)\dot{\phi}$. But then my definition in my book says that the Hamilton is given by:
$$H(\theta, \phi, \dot{\theta}, \dot{\phi})=v(\theta, \phi, \dot{\theta}, \dot{\phi})p_{\phi}-L(\theta, v(\theta, \phi, \dot{\theta}, \phi), \dot{\theta}, \dot{\phi})$$ where $v(\theta, \phi, \dot{\theta}, \dot{\phi})$ is determined as the unique solution to the equations: $$p_i=\frac{\partial L}{\partial v_i} (\theta, \phi, \dot{\theta}, \dot{\phi})$$ But I'm not sure how to find the $v(\theta, \phi, \dot{\theta}, \dot{\phi})$, do I have to solve more than one equation?
Your equation for Hamiltonian is slightly off. For a Lagrangian $L(\mathbf{q}, \dot {\mathbf q})$, the Hamiltonian is $$H(\mathbf{q}, \dot {\mathbf q})=\sum_i p_i\dot{q_i}-L(\mathbf{q}, \dot {\mathbf q})$$ In your case $q_1=\theta$ and $q_2=\phi$. So you need to calculate $$p_\theta=\frac{\partial L(\mathbf{q}, \dot {\mathbf q})}{\partial \dot{\theta}}\\p_\phi=\frac{\partial L(\mathbf{q}, \dot {\mathbf q})}{\partial \dot{\phi}}$$ Therefore you get $$H(\theta,\phi,\dot{\theta},\dot{\phi})=\left[\dot{\theta}\frac{\partial}{\partial\dot\theta}+\dot\phi\frac{\partial}{\partial\dot\phi}-1\right]L(\theta,\phi,\dot\theta,\dot\phi)$$