Two things fix the answer: the correct form of the statistic, and the correct degrees of freedom.

Form. Pearson's statistic divides by the expected frequency:

$\chi^{2}=\sum_{i=1}^{k}\frac{(o_i-e_i)^{2}}{e_i},\qquad e_i=n\,p_{0i}.$

Option (D) divides by $o_i$ , which is wrong (and undefined for an empty cell).

Degrees of freedom. There are $k$ cells, but the counts satisfy the single linear constraint $\sum_i o_i=n$ , so only $k-1$ of the deviations are free. Since $\uline{p}_0$ is completely specified, no parameters are estimated from the data and nothing further is subtracted. Hence

$\sum_{i=1}^{k}\frac{(o_i-e_i)^{2}}{e_i} \overset{d}{\longrightarrow} \chi^{2}_{k-1}.$

That is option (B), and it rules out (C).

Option (A) is wrong on two counts: the degrees of freedom for a multinomial goodness-of-fit test do not involve the sample size $n$ , and the likelihood ratio statistic $2\sum o_i\log(o_i/e_i)$ is also asymptotically $\chi^{2}_{k-1}$ , being asymptotically equivalent to Pearson's.

(The rule in general: $\mathrm{df}=k-1-(\text{number of parameters estimated})$ ; here that last term is $0$ .)