This is conceptually very simple, so the only "difficult" part will be the calculations: if #n# is the number of variabiles of a function (in your case thus #n=2#, then the gradient of the function is a vector of length #n#, whose elements are the derivatives with respect to each variables, so if you have #f(x_1,...,x_n)#, the gradient will be the vector
#({\partial f}/{\partial x_1},..,{\partial f}/{\partial x_n})#
So, the gradient of your function #p# will be the 2-dimensional vector
#({\partial p}/{\partial x},{\partial p}/{\partial y})#.
Let's calculate the two derivatives: by the chain rule, if we need to derive an expression such #\sqrt{f(x,y)}#, the derivative will be #{f'(x,y)}/{2\sqrt{f(x,y)}}#, where of course #f'(x,y)# is the derivative with respect to the right variable. So, deriving with respect to #x# (which means that we must consider #y# as a constant) , we get #f'(x,y)=-8x#.
Deriving with respect to #y#, we get instead #-2y#. So, the gradient vector will be
#( {-8x}/{2\sqrt{24-4x^2-y^2}}, {-2y}/{2\sqrt{24-4x^2-y^2}})#
We can simplify both terms, obtaining
#( {-4x}/{\sqrt{24-4x^2-y^2}}, {-y}/{\sqrt{24-4x^2-y^2}})#
To evaluate this vector, we only need to plug in the values, if #x=-2# and #y=1#, we get
#(8/sqrt(7), -1/sqrt(7) )#
