= Solution
Translation symmetry in the plane and reflection symmetry across it force the <magnetic field> to be uniform on each side, parallel to $\mathbf e_y$, and opposite on the two sides. Apply <Ampère's law> to a rectangular loop perpendicular to the sheet whose two long sides have length $L$. The enclosed current is $KL$, so
$$
2BL=\mu_0KL.
$$
The right-hand rule fixes the directions:
$$
\boxed{
\mathbf B(z>0)=-\frac{\mu_0K}{2}\mathbf e_y,
\qquad
\mathbf B(z<0)=\frac{\mu_0K}{2}\mathbf e_y}.
$$
Consequently
$$
\mathbf e_z\times\mathbf B(0^+)
-\mathbf e_z\times\mathbf B(0^-)
=\frac{\mu_0K}{2}\mathbf e_x
-\left(-\frac{\mu_0K}{2}\mathbf e_x\right)
=\boxed{\mu_0\mathbf K},
$$
which is the magnetic jump condition for a <surface current density>.
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