Answer: C

Video solution:

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Written solution:

After the first $8$ passengers sit, there are $4$ empty seats among the $12$ seats, so there are $\binom{12}{4}=495$ equally likely sets of empty seats.

Count the complement, where no row has two adjacent empty seats. In a row of three seats, the possible empty-seat patterns with no adjacent empty seats have sizes $0,1,2$, with $1,3,1$ choices respectively. So we need the coefficient of $x^4$ in $$(1+3x+x^2)^4.$$ This coefficient is $$\begin{gathered} 3^4+4\cdot3\cdot3^2+\binom42 \\ =81+108+6 \\ =195. \end{gathered}$$

Therefore the probability that at least one row has adjacent empty seats is $$\frac{495-195}{495}=\frac{20}{33}.$$

Thus, C is the correct answer.