= Solution
An <unbiased estimator> $\widehat\theta$ satisfies $\mathbb E_\theta\widehat\theta=\theta$ for every parameter value. Under the standard full-column-rank assumption on the <design matrix>, the Gaussian <log-likelihood> for the <normal linear model>, up to terms independent of $\beta$, is $-\|Y-X\beta\|^2/(2\sigma^2)$. Its maximum is the <ordinary least squares> minimum. Differentiating gives $X^TX\widehat\beta=X^TY$, and full column rank makes $X^TX$ positive definite. Thus
$$
\boxed{\widehat\beta=(X^TX)^{-1}X^TY.}
$$
Since $\mathbb EY=X\beta$, its expectation is $(X^TX)^{-1}X^TX\beta=\beta$, proving \b[unbiasedness]. Estimating an unknown $\sigma^2$ as well does not change the maximizing coefficient vector.
The printed hypotheses give $p<n$ but do not explicitly require full column rank. Without that condition the <maximum-likelihood estimators> are not unique: they are $X^+Y+z$ with $z\in\ker X$, as in <rank-deficient ordinary least squares>. If $0\ne z\in\ker X$, the observation distributions at $\beta$ and $\beta+z$ are identical. Every estimator then has the same expectation at both parameters, so no estimator can be unbiased for both entire coefficient vectors. This proves <nonidentifiability prevents unbiased coefficient estimation>. The stated estimator and unbiasedness claim require the usual full-rank condition.
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