Poisson processes and coin flips

$$N\sim\mathrm{Poisson}(\lambda t).$$

$$ \Pr( N\text{ is even}) = e^{-\lambda t} \sum_{n=0}^\infty \frac{(\lambda t)^{2n}}{(2n)!}= e^{-\lambda t}\cosh(\lambda t)\to\frac 1 2\text{ as }t\to\infty. $$

OK, I misconstrued the word "flipped" (see comments below). And I also treated it as a "fair" coin above. I'll edit further, unless I don't.

Construing "flip" in the more usual sense (not just turning the coin upside down manually, but randomly getting "heads" or "tails"), the problem is simpler: The conditional probability given $N>0$ is $p$, and the conditional probability given $N=0$ is $1$. So we get \begin{align} & 1\cdot\Pr(N=0) + p \Pr(N>0) \\[8pt] = {} & e^{-\lambda t} + p (1-e^{-\lambda t}) \\[8pt] = {} & (1-p)e^{-\lambda t} + p\to p\text{ as }t\to\infty. \end{align}

The process is indeed renewed at each flip, but what you randomly generate at each renewal is the exponentially distributed time $T$ until the next renewal, satisfying $$ \Pr(T>t) = \Pr(N= 0) = e^{-\lambda t} $$ since the event $T>t$ is the same as the event $N=0$ (i.e. not just the probabilities, but the events themselves are the same, in the sense that either event occurs if and only if the other does). Here I'm construing $N$ as the number of later renewals between the time of renewal and and that time plus $t$.