For three events, $$A$$, $$B$$ and $$C$$, $$P$$(Exactly one of $$A$$ or $$B$$ occurs) = $$P$$(Exactly one of $$B$$ or $$C$$ occurs) = $$P$$(Exactly one of $$C$$ or $$A$$ occurs) = $$\frac{1}{4}$$ and $$P$$(All the three events occur simultaneously) = $$\frac{1}{16}$$. Then the probability that at least one of the events occurs, is:

Let us denote the single-event probabilities as $$P(A),\;P(B),\;P(C)$$ and the pairwise intersection probabilities as $$P(A\cap B),\;P(B\cap C),\;P(C\cap A)$$. For ease of writing we put

$$x=P(A)+P(B)+P(C),\qquad y=P(A\cap B)+P(B\cap C)+P(C\cap A).$$

First recall the standard result for two events:

Probability that exactly one of the two events $$A$$ or $$B$$ occurs is given by

$$P(\text{exactly one of }A\text{ or }B)=P(A)+P(B)-2P(A\cap B).$$

We now apply this formula to the three statements given in the question.

For $$A$$ and $$B$$ we have

$$P(A)+P(B)-2P(A\cap B)=\frac14.$$ For $$B$$ and $$C$$ we have

$$P(B)+P(C)-2P(B\cap C)=\frac14.$$ For $$C$$ and $$A$$ we have

$$P(C)+P(A)-2P(C\cap A)=\frac14.$$

Adding these three equations term-by-term, we obtain

$$\bigl[P(A)+P(B)+P(C)\bigr]+\bigl[P(A)+P(B)+P(C)\bigr]\; -\;2\bigl[P(A\cap B)+P(B\cap C)+P(C\cap A)\bigr] =\frac14+\frac14+\frac14.$$

That is

$$2\bigl[x-y\bigr]=\frac34\quad\Longrightarrow\quad x-y=\frac38.$$ Hence

$$x-y=\frac38. \quad -(1)$$

Now use the inclusion-exclusion formula for three events:

$$P(A\cup B\cup C)=x-y+P(A\cap B\cap C).$$

We are given $$P(A\cap B\cap C)=\dfrac1{16},$$ so substituting this and the value from (1) gives

$$P(A\cup B\cup C)=\frac38+\frac1{16} =\frac{6}{16}+\frac{1}{16} =\frac{7}{16}.$$

Thus the probability that at least one of the three events occurs equals $$\dfrac{7}{16}.$$

Hence, the correct answer is Option B.