If the series limit frequency of the Lyman series is $$V_L$$, then the series limit frequency of the Pfund series is:

For hydrogen-like atoms, the frequency of any spectral line is given by the Rydberg relation

$$\nu \;=\; R_{\!H}c\left(\dfrac{1}{n_f^{\,2}}-\dfrac{1}{n_i^{\,2}}\right)$$

where $$R_{\!H}$$ is the Rydberg constant for hydrogen, $$c$$ is the speed of light, $$n_f$$ is the principal quantum number of the lower (final) orbit, and $$n_i$$ is that of the higher (initial) orbit with $$n_i > n_f$$.

Now, the series limit of any spectral series corresponds to the transition in which the electron starts from $$n_i=\infty$$ and falls to the fixed lower level $$n_f$$ of that series. In that case $$\dfrac{1}{n_i^{\,2}}\to 0$$, so the formula simplifies to

$$\nu_{\text{limit}} \;=\; R_{\!H}c\left(\dfrac{1}{n_f^{\,2}}-0\right) \;=\; \dfrac{R_{\!H}c}{n_f^{\,2}}.$$

We are first told that the series limit frequency of the Lyman series is $$V_L$$. The Lyman series has $$n_f = 1$$. Substituting $$n_f = 1$$ in the above simplified expression, we indeed get

$$V_L \;=\; \dfrac{R_{\!H}c}{1^{2}} \;=\; R_{\!H}c.$$

Next, we need the series limit frequency of the Pfund series. The Pfund series corresponds to transitions that end at $$n_f = 5$$. Using the same formula with $$n_f = 5$$, we obtain

$$\nu_{\text{Pfund\;limit}} \;=\; \dfrac{R_{\!H}c}{5^{2}} \;=\; \dfrac{R_{\!H}c}{25}.$$

But from the earlier step we already have $$R_{\!H}c = V_L$$, so substituting this value we get

$$\nu_{\text{Pfund\;limit}} \;=\; \dfrac{V_L}{25}.$$

Hence, the correct answer is Option A.