= Solution
Take the square projection cell $0\leq x,y\leq1$ and label the heights rather than erasing them in projection. The Fe positions appear at the cell corners and at $(1/2,1/2)$, both with $z=0$. The Se projections are $(1/2,0)$ at height $+\eta$ and $(0,1/2)$ at height $-\eta$, together with their periodic copies on opposite edges. These are distinct height classes even though the observer looks down the $z$ axis.
The corner Fe sites together count as one atom per cell, the central Fe as one, and each edge-shared Se projection contributes half at each of two equivalent edges. The count is therefore two Fe and two Se per <primitive tetragonal lattice> cell. The Fe sublattice has its shorter nearest-neighbor directions along the projected cell diagonals. The distinction between those directions and the conventional tetragonal axes matters for the distortion in <1/d>.
\Image[/past-exam-of-the-condensed-matter-physics-major-option-course-of-the-university-of-oxford/2015-fe-se-projection-and-orthorhombic-distortion.png]
{title=FeSe projection with selenium heights and the diagonal orthorhombic distortion}
{description=Original coordinate-based projection and schematic distortion. Opposite selenium heights remain distinguished; the orthorhombic easy axes are rotated by 45 degrees from the conventional tetragonal axes. The distortion amplitude is exaggerated.}
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