It is a standard exercise to show that this is a Poisson process.
Let $X_t$ be the number of "potential" arrivals before time $t$ and let $Y_t$ be the number of actual arrivals before time $t$. Suppose $0. Then $ \begin{align} & {}\quad\Pr(Y_t-Y_s = y) \\[12pt] & {} = \Pr\left( \bigcup_{w=y}^\infty \Big(\left(X_t-X_s = w\right) \cap (y\text{ of the }w\text{ potential arrivals are actual arrivals})\Big) \right) \\[12pt] & {} = \sum_{w=y}^\infty \Pr((X_t-X_s=w) \cap (y\text{ of the }w\text{ potential arrivals are actual arrivals})) \\[12pt] & = \sum_{w=y}^\infty \frac{e^{-L(t-s)} (L(t-s))^w}{w!} \cdot \binom w y B^y (1-B)^{w-y} = \sum_{w=y}^\infty \frac{e^{-L(t-s)} (L(t-s))^w B^y (1-B)^{w-y}}{(w-y)!y!} \\[12pt] & = e^{-L(t-s)} \frac{B^y}{y!} (L(t-s))^y \sum_{w=y}^\infty \frac{(L(t-s))^{w-y}(1-B)^{w-y}}{(w-y)!} \\[12pt] & = e^{-L(t-s)} \frac{B^y}{y!} (L(t-s))^y \sum_{v=0}^\infty \frac{(L(t-s))^v (1-B)^v}{v!} \qquad\text{where } v=w-y \\[12pt] & = e^{-L(t-s)} \frac{B^y}{y!} (L(t-s))^y \cdot e^{L(t-s)(1-B)} = e^{-BL(t-s)} \frac{(BL(t-s))^y}{y!}. \end{align} $ So you have a Poisson process with rate $BL$.