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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 bottom-boundary harmonic measure of a strip isThis is the unconditional exit density on the bottom edge, not the density conditional on using that edge. Its total mass isIndeed substitution $v=e^{a-x}$ gives the positive-half-axis mass of the Poisson kernel, equal to $\frac{1}{2}+{\pi }^{-1}\arctan (\cot y)=1-y/\pi$ 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.