Well, the Dirac delta function $\delta(x)$ is a distribution, also known as a generalized function.
One can e.g. represent $\delta(x)$ as a limit of a rectangular peak with unit area, width $\epsilon$, and height $1/\epsilon$; i.e.
$$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x), $$
$$\tag{2} \delta_{\epsilon}(x)~:=~\frac{1}{\epsilon} \theta(\frac{\epsilon}{2}-|x|)
~=~\left\{ \begin{array}{ccc} \frac{1}{\epsilon}&\text{for}& |x|<\frac{\epsilon}{2}, \\
\frac{1}{2\epsilon}&\text{for}& |x|=\frac{\epsilon}{2}, \\
0&\text{for} & |x|>\frac{\epsilon}{2}, \end{array} \right. $$
where $\theta$ denotes the Heaviside step function with $\theta(0)=\frac{1}{2}$.
The product $\delta(x)^2$ of the two Dirac delta distributions does strictly speaking not$^1$ make mathematical sense, but for physical purposes, let us try to evaluate the integral of the square of the regularized delta function
$$\tag{3} \int_{\mathbb{R}}\! dx ~\delta_{\epsilon}(x)^2
~=~\epsilon\cdot\frac{1}{\epsilon}\cdot\frac{1}{\epsilon}
~=~\frac{1}{\epsilon} ~\to~ \infty
\quad \text{for} \quad \epsilon~\to~ 0^+. $$
The limit is infinite as Griffiths claims.
It should be stressed that in the conventional mathematical theory of distributions, the eq. (2.95) is a priori only defined if $f$ is a smooth test-function. In particular, it is not mathematically rigorous to use eq. (2.95) (with $f$ substituted with a distribution) to justify the meaning of the integral of the square of the Dirac delta distribution. Needless to say that if one blindly inserts distributions in formulas for smooth functions, it is easy to arrive at all kinds of contradictions! For instance,
$$ \frac{1}{3}~=~ \left[\frac{\theta(x)^3}{3}\right]^{x=\infty}_{x=-\infty}~=~\int_{\mathbb{R}} \!dx \frac{d}{dx} \frac{\theta(x)^3}{3} $$ $$\tag{4} ~=~\int_{\mathbb{R}} \!dx ~ \theta(x)^2\delta(x)
~\stackrel{(2.95)}=~ \theta(0)^2~=~\frac{1}{4}.\qquad \text{(Wrong!)} $$
--
$^1$ We ignore Colombeau theory. See also this mathoverflow post.
Consider making the substitution $k = p/\hbar$ in the first expression, while simultaneously defining
$$
\sqrt{\hbar} \,a(p) = \phi(k)
$$
then the first integral will become
$$
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \sqrt{\hbar}\, a(p) e^{i\left(\frac{p}{\hbar} x - \frac{\hbar}{2m}\frac{p^2}{\hbar^2}t\right)} \frac{dp}{\hbar}
$$
which is exactly the second integral. It's just a change of notation!
Best Answer
Let's accept the last step in your "derivation". A squared delta already is an hint of serious problems.
Simply put, the expression $\delta(0)$ makes little sense. Consider the $\delta(x)$ "function" (it is a distribution, but I'll go with the simplest answer possible, just for intuition) and a succession approximating it:
$$ \delta(x-\lambda)=\lim_{\epsilon\to0}\frac{1}{\sqrt{2\pi}\epsilon}e^{\frac{(x-\lambda)^2}{2\epsilon^2}} $$ Now, you can see that, if you "evaluate" this limit for any $x$ that is non $\lambda$, you get 0: the exponential is always stronger than the power in the denominator, and dictates the limit.
But, if you take $x=\lambda$, as in your case, the exponential disappears: nothing stops the denominator from bringing the whole limit to infinity. So, you can say that $\delta(\lambda-\lambda)=\delta(0)=\infty$. So, the position eigenstate does not have a norm.