= Solution
Use the open strip $S=\{x+iy:0<y<\pi\}$; the printed closed strip denotes the region together with its exit boundary. Starting at $z=x+iy$, the <conformal bijection> $e^z$ sends $S$ onto the <complex upper half-plane> and the bottom edge $a\in\mathbb R$ onto $v=e^a>0$. The upper edge maps to the negative half-axis. The transformed initial point is $e^x\cos y+i e^x\sin y$.
By <conformal invariance of planar Brownian motion>, transform the given <Poisson kernel for the upper half-plane> and include the <Jacobian determinant> $dv/da=e^a$. The \b[<bottom-boundary harmonic measure of a strip>] is
$$
\boxed{\rho_z(a)=\frac{e^{a+x}\sin y}{\pi\bigl((e^a-e^x\cos y)^2+e^{2x}\sin^2y\bigr)}=\frac{\sin y}{2\pi\bigl(\cosh(a-x)-\cos y\bigr)}.}
$$
This is the unconditional exit density on the bottom edge, not the density conditional on using that edge. Its total mass is
$$
\boxed{\int_{\mathbb R}\rho_z(a)da=1-\frac y\pi.}
$$
Indeed substitution $v=e^{a-x}$ gives the positive-half-axis mass of the <Poisson kernel>, equal to $\frac12+\pi^{-1}\arctan(\cot y)=1-y/\pi$ for $0<y<\pi$. To obtain the conditional bottom-edge density, divide by $1-y/\pi$. The top-edge density is $\sin y/[2\pi(\cosh(a-x)+\cos y)]$ and has mass $y/\pi$, a useful normalization check.
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