# These two definitions of a hyperplane are equivalent?

My book first defines a hyper plane in $R^n$ as set $H= \{p^tx=\alpha \}$ where $p$ is a nonzero vector and $\alpha$ is scalar, or equivalently as the set $H=\{x:p^t(x-\bar{x })=0$.

Next it states that the set defined as $\{(x,y): y=f(\bar {x } )+ \xi^t(x-\bar {x } )$ is a hyperplane.

But how can this later set be written in either of the first two forms?

The equation $y=f(\bar {x } )+ \xi^t(x-\bar {x } )$ can be rewritten $y=\xi^t(x-\bar {x } )+f(\bar {x } )-y=0$, which has the form $$(\xi,-1)^t\bigl((x,y)-(\bar x,f(\bar x)\bigr)=0.$$ (You will need to read $(a,b)$ above as a column vector with $a$ above $b$.)
$\newcommand{\R}{\mathbf{R}}$Your book's assertion is correct. To see why, note that the pair $X := (x, y)$, which may be viewed as an element of $\R^{n+1}$ in the second instance, plays the role of $x$ in the definition. Setting $$\Xi = (\xi, -1),\quad \bar{X} = \bigl(\bar{x}, f(\bar{x})\bigr),$$ the condition $$y = f(\bar{x}) + \xi^{t}(x - \bar{x})$$ can be written $$\Xi^{t}(X - \bar{X}) = 0.$$
Your two conditions are not equivalent, since not every non-zero vector in $\R^{n+1}$ has a non-zero final component. That is, the "graph-like" definition cannot accommodate hyperplanes whose normal vector has final component equal to zero. (Analogously, every line in the plane is the solution set of some equation $ax + by = c$, and every equation $y = b + mx$ defines a line, but not every line can be written in slope-intercept form.)