Weinberg's normalization convention for momentum eigenstates

In this answer https://physics.stackexchange.com/a/376193/274751 two different conventions for the normalization of momentum eigenstates are mentioned. This convention amounts to the choice of $$N(p)$$ in the following definition of a momentum eigenstate (Weinberg, p. 64, eq. 2.5.5):

$$|p \rangle = N(p) U(L(p))| k \rangle$$

The inner product between momentum eigenstates, in terms of this $$N(p)$$, becomes (Weinberg, p. 67, eq. 2.5.17):

$$\langle p | p' \rangle = |N(p)|^2 \left(\frac{p_0}{k_0}\right) \delta^3(\vec{p}-\vec{p}')$$

Weinberg chooses $$N(p) = \sqrt{\frac{k_0}{p_0}}$$, so that

$$\langle p'| p \rangle = \delta^3(\vec{p}-\vec{p}')$$

The other convention is the one mentioned in the linked answer, and in it, we have

$$\langle p'| p \rangle = 2p_0 \delta^3(\vec{p}-\vec{p}')$$

My question is what would be the $$N(p)$$, as defined by Weinberg, that is used for this second convention? From a simple comparison with eq. 2.5.17 we have that $$N(p) = \sqrt{2k_0} = \sqrt{2mc}$$, but this doesn't look right. Why does it not depend on $$p$$? Are there any further differences between the treatments that should be taken into account here?

• It seems that you have your answer already. Why doesn't it look right? Who said $N(p)$ can't be constant with respect to $p$? Aug 25, 2021 at 23:14