The correct option is:
$-\ln 2$

To find the solution, we begin with the given first-order ordinary differential equation:
$(1+\sin x)\frac{dy}{dx}+\cos x=0$

We can rewrite this equation to separate the variables $y$ and $x$:
$(1+\sin x)\frac{dy}{dx}=-\cos x$
Dividing both sides by $1+\sin x$ gives:
$\frac{dy}{dx}=-\frac{\cos x}{1+\sin x}$

Now, we integrate both sides with respect to $x$:
$\int dy=-\int \frac{\cos x}{1+\sin x}dx$

To evaluate the integral on the right-hand side, we use the substitution method. Let:
$u=1+\sin x$
Then, the differential of $u$ is:
$du=\cos xdx$

Substituting these into the integration equation:
$y=-\int \frac{1}{u}du$
$y=-\ln \left|u\right|+C$
Substituting back $u=1+\sin x$:
$y=-\ln (1+\sin x)+C$

We are given the initial condition $f\left(0\right)=0$, which means $y=0$ when $x=0$. We substitute these values into our general solution to find the integration constant $C$:
$0=-\ln (1+\sin 0)+C$
Since $\sin 0=0$:
$0=-\ln \left(1\right)+C$
Since $\ln 1=0$:
$C=0$

Therefore, the particular solution is:
$f\left(x\right)=-\ln (1+\sin x)$

Now, we calculate $f\left(\frac{\pi }{2}\right)$:
$f\left(\frac{\pi }{2}\right)=-\ln (1+\sin \frac{\pi }{2})$
Since $\sin \frac{\pi }{2}=1$:
$f\left(\frac{\pi }{2}\right)=-\ln (1+1)=-\ln 2$