Answer: The moment of inertia is \(\dfrac23Ma^2(7k+20)\), and \(\boxed{k=4}\).
(i) Start with the square lamina. Its mass is \(kM\) and its side length is \(a\). The moment of inertia of a square lamina about an axis through its centre perpendicular to its plane is
\[I_{\text{centre}}=\frac16(kM)a^2.\]
The centre of the lamina is \(\frac32a\) horizontally and \(\frac32a\) vertically from \(A\). Hence its distance squared from \(A\) is
\[\left(\frac32a\right)^2+\left(\frac32a\right)^2=\frac92a^2.\]
By the parallel axis theorem, the lamina's moment of inertia about the axis through \(A\) is
\[I_{\text{lamina}}=\frac16kMa^2+kM\left(\frac92a^2\right)=\frac{14}{3}kMa^2.\]
Now consider the frame. The rods \(AB\) and \(AD\) each have length \(2a\), mass \(M\), and one end at \(A\). Therefore each contributes
\[\frac13M(2a)^2=\frac43Ma^2.\]
The rods \(BC\) and \(CD\) are each a distance from \(A\). For either of these rods, the moment of inertia about its own centre is
\[\frac1{12}M(2a)^2=\frac13Ma^2.\]
The centre of each of these two rods is distance squared \(5a^2\) from \(A\), so each contributes
\[\frac13Ma^2+5Ma^2=\frac{16}{3}Ma^2.\]
Thus the moment of inertia of the frame is
\[2\left(\frac43Ma^2\right)+2\left(\frac{16}{3}Ma^2\right)=\frac{40}{3}Ma^2.\]
So the total moment of inertia of the object is
\[I=\frac{14}{3}kMa^2+\frac{40}{3}Ma^2.\]
Factorising gives
\[I=\frac23Ma^2(7k+20),\]
as required.
(ii) Use conservation of energy. When \(D\) is first vertically below \(A\), the object has effectively turned through \(180^\circ\).
The centre of the lamina falls by \(3a\), so the lamina loses gravitational potential energy
\[kMg(3a).\]
For the four rods of the frame, the total loss of gravitational potential energy is
\[8Mga.\]
Hence the total loss of gravitational potential energy is
\[Mga(3k+8).\]
This becomes rotational kinetic energy:
\[\frac12I\omega^2=Mga(3k+8).\]
Substitute \(I=\frac23Ma^2(7k+20)\):
\[\frac12\cdot\frac23Ma^2(7k+20)\omega^2=Mga(3k+8).\]
So
\[\omega^2=\frac{3g(3k+8)}{a(7k+20)}.\]
The given angular speed is
\[\omega=\frac12\sqrt{\frac{5g}{a}},\]
so
\[\omega^2=\frac{5g}{4a}.\]
Equate the two expressions:
\[\frac{3g(3k+8)}{a(7k+20)}=\frac{5g}{4a}.\]
Cancel \(g/a\):
\[\frac{3(3k+8)}{7k+20}=\frac54.\]
Therefore
\[12(3k+8)=5(7k+20).\]
So
\[36k+96=35k+100,\]
and hence
\[\boxed{k=4}.\]
