= Solution
On the axis, the <Poisson kernel for the upper half-space> depends only on the radial boundary coordinate. Integrating over the disk gives
$$
\phi(0,0,z)=\frac z{2\pi}\int_0^{2\pi}d\theta\int_0^a\frac{r\,dr}{(r^2+z^2)^{3/2}}=z\left[-\frac1{\sqrt{r^2+z^2}}\right]_0^a.
$$
Therefore the <axial Poisson extension of a circular boundary patch> is
$$
\boxed{\phi(0,0,z)=1-\frac{z}{\sqrt{a^2+z^2}},\qquad z>0.}
$$
It tends to one as the axis approaches the boundary inside the patch, and tends to zero as $a^2/(2z^2)$ far above it.
The boundary values jump at the circular rim. They are understood almost everywhere, or as limits at continuity points; changing the assigned value on that measure-zero circle does not change the integral. At a rim point the vertical limit is one-half, rather than a continuous extension with the specified value on both sides of the jump.
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