To determine which of the given compounds have all atoms in one plane in all possible conformations, we need to analyze the geometry and conformational flexibility of each molecule.

Step 1: Analyze Option A (Biphenyl)

1. Structure: Biphenyl consists of two benzene rings connected by a carbon-carbon single bond.
2. Planarity of Rings: Each benzene ring is individually planar because all its carbon atoms are $sp^2$ hybridized.
3. Conformations: There is free rotation around the C-C single bond connecting the two rings. This rotation leads to different conformations.
4. Steric Hindrance: In the gas phase, the most stable conformation is one where the two rings are twisted relative to each other by about 44 degrees. This twisted conformation is non-planar. The reason for this twist is to minimize the steric repulsion between the hydrogen atoms on the ortho positions of the two rings.
5. Conclusion: Since biphenyl exists in non-planar conformations, not all of its atoms lie in a single plane in all possible conformations. Therefore, option A is incorrect.

Step 2: Analyze Option B (Naphthalene)

1. Structure: Naphthalene is an aromatic hydrocarbon consisting of two fused benzene rings.
2. Hybridization: All ten carbon atoms in the naphthalene molecule are $sp^2$ hybridized.
3. Rigidity and Planarity: The fused ring system makes the molecule rigid. There are no single bonds around which significant rotation can occur, so there is essentially only one conformation.
4. Geometry: Due to the $sp^2$ hybridization of all carbon atoms and the fused ring structure, the entire molecule, including all carbon and hydrogen atoms, is planar.
5. Conclusion: Since naphthalene has a single, rigid, planar conformation, all its atoms are in one plane. Therefore, option B is correct.

Step 3: Analyze Option C (Ketene, $H_2C = C = O$)

1. Structure: Ketene has the structure $H_2C_a=C_b=O$.
2. Hybridization:
   • The central carbon atom ($C_b$) is involved in two double bonds, so it is $sp$ hybridized.
   • The terminal carbon atom ($C_a$) is involved in one double bond and two single bonds, so it is $sp^2$ hybridized.
3. Geometry:
   • The $sp$ hybridization of the central carbon ($C_b$) means that the atoms bonded to it ($C_a$ and O) are arranged linearly. Thus, the $C_a-C_b-O$ skeleton is linear.
   • The $sp^2$ hybridization of the terminal carbon ($C_a$) means that the two hydrogen atoms and the central carbon atom ($C_b$) attached to it lie in the same plane.
   • Since the $H_2C_a$ group is planar, and the oxygen atom is collinear with the $C_a-C_b$ bond, the entire molecule must be planar.
4. Conformations: The presence of double bonds restricts rotation, making the molecule rigid. It exists in only one conformation.
5. Conclusion: Since the single conformation of ketene is planar, all its atoms are in one plane. Therefore, option C is correct.

Step 4: Analyze Option D (Allene, $H_2C = C = CH_2$)

1. Structure: Allene has the structure $H_2C_a=C_b=C_cH_2$.
2. Hybridization:
   • The central carbon atom ($C_b$) is $sp$ hybridized.
   • The terminal carbon atoms ($C_a$ and $C_c$) are $sp^2$ hybridized.
3. Geometry:
   • The $sp$ hybridization of $C_b$ results in a linear $C_a-C_b-C_c$ skeleton.
   • The central carbon atom ($C_b$) uses two perpendicular p-orbitals (e.g., $p_y$ and $p_z$) to form the two $\pi$-bonds with the terminal carbons.
   • The plane of the $H_2C_a$ group is determined by the p-orbital used for its $\pi$-bond. If $C_b$ uses $p_y$ to bond with $C_a$, the $H_2C_a$ plane will be the xz-plane.
   • Consequently, $C_b$ must use its other p-orbital, $p_z$, to bond with $C_c$. This forces the $H_2C_c$ plane to be the xy-plane.
   • The two terminal $CH_2$ groups lie in planes that are perpendicular to each other.
4. Conclusion: Allene is a non-planar molecule. Since its only conformation is non-planar, it does not satisfy the condition. Therefore, option D is incorrect.

Final Answer Summary:

• A: Not always planar. Incorrect.
• B: Rigid and planar. Correct.
• C: Rigid and planar. Correct.
• D: Rigid and non-planar. Incorrect.

The compounds in which all atoms are in one plane in all possible conformations are Naphthalene (B) and Ketene (C).