1. Find the electronic configurations of the given lanthanides

For lanthanides, the common configuration is based on xenon core, i.e. $[Xe]$.

• $\text{La} (Z=57): [Xe]5d^16s^2$
• $\text{Ce} (Z=58): [Xe]4f^15d^16s^2$
• $\text{Nd} (Z=60): [Xe]4f^46s^2$
• $\text{Yb} (Z=70): [Xe]4f^{14}6s^2$

2. Write the configurations in the $+2$ oxidation state

In the $+2$ state, two electrons are removed, generally from the $6s$ orbital first.

• $\text{La}^{2+}: [Xe]5d^1$
• $\text{Ce}^{2+}: [Xe]4f^15d^1$ (or effectively one unpaired electron remains in inner levels; in any case not diamagnetic)
• $\text{Nd}^{2+}: [Xe]4f^4$
• $\text{Yb}^{2+}: [Xe]4f^{14}$

3. Check diamagnetic nature

A species is diamagnetic if all electrons are paired.

• $\text{La}^{2+}$ has $5d^1$ $\Rightarrow$ paramagnetic
• $\text{Ce}^{2+}$ has unpaired electron(s) $\Rightarrow$ paramagnetic
• $\text{Nd}^{2+}$ has $4f^4$ $\Rightarrow$ paramagnetic
• $\text{Yb}^{2+}$ has $4f^{14}$, a completely filled $f$-subshell $\Rightarrow$ diamagnetic

4. Identify the lanthanide showing stable $+2$ oxidation state with diamagnetic nature

$\text{Yb}^{2+}$ is especially stable because of the fully filled $4f^{14}$ configuration, and it is diamagnetic.

Therefore, the correct option is:

$\boxed{\text{Yb}}$

So, Option B is correct.