= Solution
There are no $\partial_tN$ or $\partial_tN^i$ terms in the displayed <Einstein-Hilbert action>, so the <lapse function> and <shift vector> have zero <canonical momenta>. Their equations impose <constraint equations in field theory> rather than independent propagation; they are <Lagrange multipliers> in the <Hamiltonian formulation of general relativity>.
With the question's positive sign for $K_{ij}=(\dot h_{ij}-D_iN_j-D_jN_i)/(2N)$, vary $\dot h_{ij}$ at fixed $h,N,N^i$. Since $\delta K_{ij}=\delta\dot h_{ij}/(2N)$,
$$
\delta\mathcal L=\sqrt h\,(K^{ij}-Kh^{ij})\delta\dot h_{ij},\qquad \boxed{\pi^{ij}=\sqrt h\,(K^{ij}-Kh^{ij}).}
$$
Thus the <canonical momentum of the spatial metric> is a weight-one <tensor density>. Write $\pi=h_{ij}\pi^{ij}$; in three spatial dimensions $\pi=-2\sqrt hK$, so
$$
K^{ij}=\frac1{\sqrt h}\left(\pi^{ij}-\frac12\pi h^{ij}\right).
$$
The action has suppressed its overall $1/(16\pi G)$, and this normalization is retained here. Reversing the sign convention for <extrinsic curvature> would reverse the momentum relation; the calculation uses exactly the convention supplied in this question.
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