I am the author of the text referenced. The integral over infinite volume is indeed the delta function (times (2pi)^3) and seems in this context more straightforwardly expressed as e^2ipx. (There was a pedagogic reason for expressing it as two factors in (4-27).) In the text, the integral is actually over finite volume V and (4-27) has another factor in front of it. The integral is zero except when p = 0 (with the usual boundary conditions at the edges of V.) When p = 0, the integral is V, but the other factor is zero. Hence (4-27) is zero in any case. This could have been shown more clearly, and I am posting a comment on this on the book website.
We wish to evaluate:
$$\int_0^{2\pi} P_l^m(\cos\theta)P_{l-1}^m(\cos\theta)\,\text{d}\theta $$
We use the transformation:
$$x = \cos\theta\quad \frac{dx}{d\theta} = -\sin\theta \quad d\theta = \frac{-1}{\sqrt{1-x^2}}dx$$ to obtain$$\int_1^1 \frac{-P_l^m(x)P_{l-1}^m(x)}{\sqrt{1-x^2}}dx = 0$$
Since you're integrating from 1 to 1.
Weird? Yes I thought so too. Seeing as the integral comes from spherical harmoics, I assume you are messing up your limits of integration (should be from $0$ to $\pi$, since that is how the zenith angle is normally defined).
Then, I think you should be working with:
$$\int_0^{\pi} P_l^m(\cos\theta)P_{l-1}^m(\cos\theta)\,\text{d}\theta $$
which after the transformation becomes:
$$I = \int_{-1}^1 \frac{P_l^m(x)P_{l-1}^m(x)}{\sqrt{1-x^2}}dx$$
The Legendre Polynomials satisfy the relationship:
$$P_l^m(-x) = (-1)^{l+m}P_l^m(x)$$
This means that if $l+m$ is even (odd), the Legendre Polynomial is also even (odd). Here we have the two numbers:
$l+m$ and $l+m-1$ One of these two has to be even and the other odd. So one of the polynomials is even and the other is odd. Since $\frac{1}{\sqrt{1-x^2}}$ is even, the integrand as a whole is odd. Thus the integral is also zero.
$$I = 0$$
Reference for parity relationship.
Best Answer
According to the definitions of the associated Legendre polynomials and assuming $m+1\leq l,k$ we have that
$$\int_{-1}^1 (1-x^2)\partial_x P_l^m \partial_x P_k^m \:dx = \int_{-1}^1 P_l^{m+1}P_k^{m+1}\:dx = \frac{2(l+m+1)!}{(2l+1)(l-m-1)!}\delta_{kl}$$
the orthogonality condition for fixed $m$.