They have already defined the relevant variables, namely
$d = quantity\ sold\ (in\ thousands\ of\ pounds);$
$p = sales\ price\ per\ unit;$
$r = revenue\ (in\ thousands\ of\ pounds);$
$c = cost\ of\ production\ (in\ thousands\ of\ pounds);\ and$
$n = profit\ (in\ thousands\ of\ pounds).$
Nothing hard about that. It has already been given to you.
And they have given you a number of relationships, namely
$n = r - c;$
$r = d * p;$
$p = 16 - 2d;\ and$
$c= 8 + 3d.$
Again that is given to you. All this problem is asking you to do is first to find r in terms of d, which involves working out
$r = p * d = (16 - 2d) * d = what?$ Notice that you now have r equal to an expression in d, which just happens to be quadratic.
Then it wants you to find n in terms of d, which involves working out
$n = r - c = r - (8 + 3 * d) = what\ in\ terms\ of\ d?$
Again it just happens that the final expression in d is a quadratic.
You probably would have got this if the word "quadratic" had not been there. There is no necessity for sales, cost, revenue, or profit to be quadratics. It just works out that IN THIS CASE revenue and profit are quadratic.
Edit: By the way, this type of question should be put in the Business Math forum in the future. You will get a quicker answer.
In post #2, I explained what needs to be done.
Way back, when you were studying beginning algebra, you were undoubtedly asked to express multi-variable equations in terms of a given variable.
Example: Express y in terms of x given $x^3 + 3x^2y + 3xy^2 + y^3 = 27y^3.$
$x^3 + 3x^2y + 3xy^2 + y^3 = 27y^3\implies (x + y)^3 = (3y)^3 \implies x + y = 3y \implies 2y = x \implies y = \dfrac{x}{2}.$
This is a similar problem.