= Solution
Assume each sampled response is truthful, card selection is independent of the student's status, and responses follow the usual independent sampling model. In <randomized response>, a yes answer occurs either for a user receiving the first question or for a nonuser receiving the complementary question. Its probability is
$$
\boxed{q=xp+(1-x)(1-p)=(1-x)+(2x-1)p}.
$$
Consequently $Y\sim\operatorname{Bin}(n,q)$, and the <likelihood> is
$$
L(p)=\binom nY[(1-x)+(2x-1)p]^Y[x-(2x-1)p]^{n-Y},\qquad0\leq p\leq1.
$$
For $x\ne1/2$, first maximize the <binomial likelihood> in the unrestricted probability $q$: its maximum occurs at $Y/n$. Inverting the affine relation gives
$$
\widetilde p=\frac{Y-n(1-x)}{n(2x-1)}.
$$
For an interior value $0<Y<n$, the same result follows from the score
$$
\ell'(p)=(2x-1)\frac{Y-nq}{q(1-q)}.
$$
The endpoint cases follow by monotonicity. This <estimator> has <expected value> $p$ and <variance>
$$
\boxed{\mathbb E\widetilde p=p,\qquad\operatorname{Var}(\widetilde p)=\frac{q(1-q)}{n(2x-1)^2}=\frac{p(1-p)}n+\frac{x(1-x)}{n(2x-1)^2}}.
$$
These identities follow immediately from $\mathbb EY=nq$ and $\operatorname{Var}(Y)=nq(1-q)$.
The printed claim needs a parameter-space qualification. The admissible yes probabilities lie in $[\min(x,1-x),\max(x,1-x)]$, which may exclude $Y/n$. The constrained <maximum-likelihood estimator> is
$$
\boxed{\widehat p_{\rm ML}=\min\{1,\max\{0,\widetilde p\}\}}.
$$
The displayed untrimmed <estimator> is the <maximum-likelihood estimator> only when it already lies in $[0,1]$. Its <unbiased estimator> property does not carry over to the clipped <estimator> in general. For example, if $n=1$, $x=3/4$ and $p=0$, the clipped estimate equals 1 exactly when the answer is yes, so its expectation is $1/4$, not zero. At $x=1/2$, $q=1/2$ for every $p$: the parameter is unidentifiable and the printed quotient is undefined. Thus the unbiased formula requires $x\ne1/2$, and the constrained <likelihood> assertion requires the stated clipping.
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