= Solution
The action contains neither $\dot N$ nor $\dot N^i$, so their <conjugate momentum> vanishes. They impose constraints rather than independent evolution equations: after the <Legendre transform in mechanics>, the <lapse function> and <shift vector> are <Lagrange multipliers> in the <Hamiltonian formulation of general relativity>.
Use precisely the paper's sign convention for <extrinsic curvature>, $K_{ij}=(\dot h_{ij}-D_iN_j-D_jN_i)/(2N)$. At fixed $h_{ij},N,N^i$,
$$
\delta K_{ij}=\frac{\delta\dot h_{ij}}{2N},\qquad
\delta\bigl[\sqrt hN(K_{ab}K^{ab}-K^2)\bigr]
=\sqrt h(K^{ij}-Kh^{ij})\delta\dot h_{ij}.
$$
Consequently the <canonical momentum of the spatial metric> is
$$
\boxed{\pi^{ij}=\sqrt h(K^{ij}-Kh^{ij}).}
$$
This is a weight-one <tensor density>, with indices raised and lowered using $h_{ij}$. Its trace and inverse relation are
$$
\pi=h_{ij}\pi^{ij}=-2\sqrt hK,\qquad
K^{ij}=\frac1{\sqrt h}\left(\pi^{ij}-\frac12\pi h^{ij}\right).
$$
The sign of $K_{ij}$ used here is opposite to the convention in <extrinsic curvature of a spatial hypersurface>; mixing the two conventions would reverse the momentum formula.
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