Condition for degenerate orbitals

Electrons are in degenerate orbitals if they have the same energy.

For multi-electron atoms, the energy of an orbital depends mainly on $n$ and $l$.
So, orbitals are degenerate when they have the same $n$ and same $l$; different $m_l$ values within the same subshell are degenerate.

Also, $m_s$ does not affect orbital energy in the absence of an external magnetic field.

Check pair (A)

• Electron (a): $n=3,\ l=1,\ m_l=1,\ m_s=+\frac{1}{2}$
  This is a $3p$ orbital.
• Electron (b): $n=3,\ l=2,\ m_l=1,\ m_s=+\frac{1}{2}$
  This is a $3d$ orbital.

Since $l$ is different ($1 \neq 2$), they belong to different subshells, so they are not degenerate.

Check pair (B)

• Electron (a): $n=3,\ l=2,\ m_l=-2,\ m_s=-\frac{1}{2}$
• Electron (b): $n=3,\ l=2,\ m_l=-1,\ m_s=-\frac{1}{2}$

Here both have the same $n=3$ and $l=2$, so both are in the $3d$ subshell.
Different $m_l$ values correspond to different orbitals within the same subshell, which are degenerate.

Hence, (B) is correct.

Check pair (C)

• Electron (a): $n=4,\ l=2,\ m_l=2,\ m_s=+\frac{1}{2}$
  This is a $4d$ orbital.
• Electron (b): $n=3,\ l=2,\ m_l=2,\ m_s=+\frac{1}{2}$
  This is a $3d$ orbital.

Since $n$ is different ($4 \neq 3$), these are different subshells with different energies, so they are not degenerate.

Final conclusion

Only pair (B) consists of electrons in degenerate orbitals.

Therefore, the correct option is:

$\boxed{\text{B: Only (B)}}$