I'm struggling on the following question
My approach to the question is as follows:
The equation of line $BC$ is: $\lambda_1$b+$(1-\lambda_1)$c
The equation of line $AP$ is: $\lambda_2(\beta$b$+\gamma$c$)$, for scalars $\lambda_1, \lambda_2$
Hence the intercept of the two lines occurs when $\lambda_1$b+$(1-\lambda_1)$c$=\lambda_2(\beta$b$+\gamma$c$)$
i.e this would be the position of $D$, but then how can I go further to show that $BD:DC=\gamma : \beta$
With your method, comparing coefficients of $\mathbf b$ and $\mathbf c$, $$\begin{cases} \lambda_1 - \beta\lambda_2 = 0\\ \lambda_1 + \gamma \lambda_2 = 1 \end{cases}$$ Solving the system of linear equations, $$\lambda_2 = \frac{1}{\beta + \gamma}, \quad\lambda_1 = \frac\beta{\beta+\gamma}$$ If $\beta + \gamma \ne 0$, then $$\begin{align*} \vec {AD} &= \frac{\beta \mathbf b + \gamma \mathbf c}{\beta + \gamma}\\ \vec {BD} &= \vec {AD}-\mathbf b = \frac{\gamma }{\beta + \gamma}(\mathbf c-\mathbf b)\\ \vec {DC} &= \mathbf c - \vec {AD} = \frac{\beta}{\beta + \gamma}(\mathbf c-\mathbf b)\\ \end{align*}$$