Imagine that you play the following game. You flip a coin that has probability $p$ of landing heads, and probability $1-p$ of landing tails.
If the result is head, you win $0$ dollars. If the result is tail, then you play a Poisson game with parameter $\lambda$, that is, a game in which you win $k$ dollars with probability $\frac{e^{-\lambda}\lambda^k}{k!}$.
Let random variable $Y$ denote your winnings in the game described above. We want to find the probability distribution of $Y$.
Let's first deal with $P(Y=0)$. You can win $0$ dollars in two ways: (i) the coin flip results in head or (ii) the coin flip results in tail, but the Poisson distribution subgame results in $0$. The probability of (i) is $p$. To find the probability of (ii), note that we must get tail (probability $1-p$) and the Poisson subgame must give result $0$ (probability $\frac{e^{-\lambda}\lambda^0}{0!}$, or more simply $e^{-\lambda}$). Thus $P(Y=0)=p+(1-p)e^{-\lambda}.$
Next we deal with $P(Y=k)$ where $k\gt 0$. To get $k$ dollars, we must have gotten tail on the coin toss (probability $1-p$ and gotten result $k$ in the Poisson subgame (probability $\frac{e^{-\lambda}\lambda^k}{k!}$). So if $k \gt 0$, then $P(Y=k)=(1-p)\frac{e^{-\lambda}\lambda^k}{k!}.$