Denote $ \angle{PAB} = \angle{PBC} = \angle{PCA} = \omega$. From $ BPCA_1 -$cyclic $ \Longrightarrow$ $ \angle AA_1C = \angle PBC = \omega$. Hence $ CA_1||AB$ $ \Longrightarrow$ $ \frac {S_{A_1BC}}{S_{ABC}} = \frac {AM}{MA_1} = \frac {MC}{MB} = \frac {S_{PAC}}{S_{PAB}}$, where $ M = BC\cap AA_1$. Hence $ S_{A_1BC} = \frac {S_{PAC}}{S_{PAB}}.S_{ABC} = \frac {PA.PC.sinA}{PA.PB.sinB}.S_{ABC} = \frac {sin\omega .sinA}{sin(C - \omega).sinB}.S_{ABC}$ $ = \frac {sin^2A}{sin^2C}.S_{ABC} = \frac {a^2}{c^2}.S_{ABC}$ (since $ cot\omega = cotA + cotB + cotC$). So the problem reduces to $ \boxed{\frac {a^2}{c^2} + \frac {c^2}{b^2} + \frac {b^2}{a^2}\geq \lambda}$. It's easy to see that $ \lambda\geq 3$ Let $ c = max\{ a,b,c\}$. If we fix $ a = b$ and assume $ f(c) = \frac {a^2}{c^2} + \frac {c^2}{a^2} + 1$ then $ f'(c) > 0$ and $ f(c)$ is continuous and if we take $ c\to a$ we get that $ \lambda = 3$

Denote $ \angle{PAB} = \angle{PBC} = \angle{PCA} = \omega$. Suppose that $AA_1$ cuts $BC$ at $A_2$ define $B_2,C_2$ similarly. Note that $$\frac {S_{A_1BC}}{S_{ABC}} = \frac{A_1A_2}{AA_2}=\frac {ABsin(B)}{A_1Bsin(A_2BA_1)}=\frac {sin(PCB)sin(B)}{sin(\omega)sin(A_2BA_1)}=\frac {BPsin(B)}{CPsin(CPA_1)} =\frac {BPsin(B)}{CPsin(A)}.$$Since the last fraction is cyclic the sum is at least $3$. Taking an equilateral triangle gives a equality case for $ \lambda = 3.$