= Solution
For $0<p<1$, the <Bernoulli distribution> has mass
$$
\Pr(Y=y)=p^y(1-p)^{1-y}
=\exp\left[y\log\frac p{1-p}+\log(1-p)\right],\qquad y\in\{0,1\}.
$$
In <exponential dispersion family> notation $\exp\{[y\theta-b(\theta)]/\phi+c(y,\phi)\}$, take
$$
\boxed{\theta=\log\frac p{1-p},\qquad b(\theta)=\log(1+e^\theta),\qquad\phi=1,\qquad c(y,1)=0.}
$$
The support is $\{0,1\}$; outside it the probability is zero. The <natural parameter of an exponential family> is the <log odds>, and the <dispersion parameter> is fixed rather than freely estimated. Differentiating the cumulant gives
$$
\boxed{\mathbb EY=b'(\theta)=\frac{e^\theta}{1+e^\theta}=p,\qquad
\operatorname{Var}(Y)=\phi b''(\theta)=\frac{e^\theta}{(1+e^\theta)^2}=p(1-p).}
$$
At $p=0$ or $1$ the degenerate distributions are limiting boundary cases with natural parameter $-\infty$ or $+\infty$.
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