This is obvious by inspection of the quantity the LASSO is optimizing.
Take the prior for $\beta_i$ to be independent Laplace with mean zero and some scale $\tau$.
So $p(\beta|\tau) \propto e^{-\frac{1}{2\tau} \sum_i|\beta_i|}$.
The model for the data is the usual regression assumption $y \stackrel{\text{iid}}{\sim}N(X\beta,\sigma^2)$.
$f(\mathbf{y}|\mathbf{X},\boldsymbol\beta,\sigma^{2}) \propto (\sigma^{2})^{-n/2} \exp\left(-\frac{1}{2{\sigma}^{2}}(\mathbf{y}- \mathbf{X} \boldsymbol\beta)^{\rm T}(\mathbf{y}- \mathbf{X} \boldsymbol\beta)\right)$
Now minus twice the log of the posterior is of the form
$k(\sigma^2,\tau,n,p)+$ $\frac{1}{{\sigma}^{2}} (\mathbf{y}- \mathbf{X} \boldsymbol\beta)^{\rm T}(\mathbf{y}- \mathbf{X} \boldsymbol\beta)+ \frac{1}{\tau} \sum_i|\beta_i|$
Let $\lambda=\sigma^2/\tau$ and we get $-2\log$-posterior of
$k(\sigma^2,\lambda,n,p)+$ $\frac{1}{{\sigma}^{2}}\left[ (\mathbf{y}- \mathbf{X} \boldsymbol\beta)^{\rm T}(\mathbf{y}- \mathbf{X} \boldsymbol\beta)+ \lambda \sum_i|\beta_i|\right]$
The MAP estimator for $\beta$ minimizes the above, which minimizes
$S=(\mathbf{y}- \mathbf{X} \boldsymbol\beta)^{\rm T}(\mathbf{y}- \mathbf{X} \boldsymbol\beta)+ \lambda \sum_i|\beta_i|$
So the MAP estimator for $\beta$ is LASSO.
(Here I treated $\sigma^2$ as effectively fixed but you can do other things with it and still get LASSO coming out.)
Edit: That's what I get for composing an answer off line; I didn't see a good answer was already posted by Andrew. Mine really doesn't do anything his doesn't do already. I'll leave mine for now because it gives a couple more details of the development in terms of $\beta$.