cis- and trans-isomers in equilibrium
Option (D) is the answer. This may be predicted by redistribution of hybrid orbital theory or simply &quot;Bent's Rule&quot;.
Clearly, during movement of the lone pair from one sp2 hybrid orbital to the other(to interconnect between the cis and trans isomers), there has to be an intermidiate transition state, during flipping when the lone pair resides in an unhybridized p-orbital. However the activation energy to achieve this intermidiate transition state is low for option (D).
Between N and H (as in the question), the electronegativity of N is higher than H. This pulls the shared electron pair in the N-H bond towards the N atom. In order to stabilize the high electron density, there is a redistribution of hybridization observed. The s character of the sp2 hybrid orbital of N (forming bond with the H ) increases ( eg from sp2 to sp1.8). This causes a corresponding decrease of s character of the other sp2 hybrid orbitals of N, such as the one containing the lone pair. The decrease in s character is associated with a a corresponding increase in p character of the lone pair containing orbital( eg from sp2 to sp2.3). Since it is easier to achieve transition from sp2.3 to unhybridized p orbital (having 100% p character), in the transition state it is easier to mediate between geometric isomer, possibly even existing in equilibrium.
For further reading see this answer:
Why are oxime geometrical isomers stable?