The electronic configurations of bivalent europium and trivalent cerium are:
(atomic number: Xe = 54, Ce = 58, Eu = 63)

We begin with the fact that xenon has the atomic number $$54$$, so its electronic configuration is taken as the core $$[Xe]$$.

Step 1 - Neutral cerium. Cerium has atomic number $$58$$, i.e. $$58-54=4$$ electrons must be placed beyond xenon. According to the Aufbau order (sub-shells are filled in increasing $$n+\ell$$ value and, for equal $$n+\ell$$, lower $$n$$ first), the order after $$6s$$ goes $$4f$$ then $$5d$$.

So, adding the four electrons one by one gives $$[Xe]6s^2\,4f^1\,5d^1.$$ This is the accepted ground-state arrangement for neutral Ce.

Step 2 - Trivalent cerium. To obtain $$\text{Ce}^{3+}$$ we must remove three electrons, and the rule is: electrons are lost first from the sub-shell having the highest principal quantum number $$n$$. Hence, the sequence of removal is $$6s$$ $$\rightarrow$$ $$5d$$ $$\rightarrow$$ $$4f$$.

Removing the two $$6s$$ electrons and the one $$5d$$ electron gives $$[Xe]4f^1.$$ Thus, $$\displaystyle \text{Ce}^{3+}: [Xe]4f^{\,1}.$$

Step 3 - Neutral europium. Europium has atomic number $$63$$, so $$63-54=9$$ electrons are to be placed after xenon. The Aufbau filling continues to load the $$4f$$ sub-shell:

$$[Xe]6s^2\,4f^7$$ provides precisely the nine additional electrons (2 in $$6s$$ and 7 in $$4f$$), giving a half-filled $$4f$$ sub-shell, which is energetically favourable. Hence neutral Eu is $$[Xe]4f^7 6s^2.$$

Step 4 - Bivalent europium. For $$\text{Eu}^{2+}$$ two electrons must be removed. Again, electrons are taken first from the highest $$n$$ level, i.e. the outer $$6s$$ sub-shell:

$$[Xe]4f^7 6s^2 \;\longrightarrow\; [Xe]4f^7.$$

Therefore, $$\displaystyle \text{Eu}^{2+}: [Xe]4f^{\,7}.$$

Step 5 - Matching with the options. Summarising, we have $$\text{Eu}^{2+}: [Xe]4f^{\,7},\qquad \text{Ce}^{3+}: [Xe]4f^{\,1}.$$ Option B lists exactly these configurations.

Hence, the correct answer is Option B.