Confusion regarding the appearance of $hbar$ in the eigenfunctions of momentum operator

Firstly it's a standard convention to set ${bf p}=hbar {bf k}$ where $|{bf k}|$ is the wavenumber and $|{bf p}|$ is the momentum. Secondly many books, and nearly all working theoretical physicsts use "natural units" in which $hbar=c=1$, so these symbols do not appear in their equations. When they need numbers in physical units (seconds, meters etc) it is easy to restore $hbar$ and $c$ by looking at the dimensions.

As for delta functions the basic identies are $$ int_{-infty}^infty frac{dk}{2pi} e^{ikx}= delta(x), $$ and $$ int_{-infty}^infty dx e^{-ikx}= 2pi delta(k). $$

The first one is correct for momentum. The second one is not given in terms of momentum, but in terms of angular wave number $k$. Note the subscript. This is a different quantity which is related to momentum by

$$p = hbar k$$

and that accounts for the presence/absence of factors $hbar$ in the expressions for the wave functions. Namely, since the above gives $k = frac{p}{hbar}$, then a simple substitution showns that

$$phi_k(x) = frac{1}{sqrt{2pi}} e^{ikx}$$

becomes, in terms of $p$,

$$phi_p(x) = frac{1}{sqrt{2pi}} e^{ileft(frac{p}{hbar}right)x}$$

which is just your first expression. You may note this is missing the $hbar$ under the square root: the reason for this is that that factor then appears if we are then trying to normalize this wave function (note here that "normalize" needs to be interpreted loosely, because these are technically not normalizable; they belong to the infinitycrust of the rigged Hilbert space, not the "proper" Hilbert space. They're delta vectors.). The normalization is defined by an integral, and when we perform substitution under an integral, we pick up some additional changes due to the differential element, here $dk$, which becomes $frac{1}{hbar} dp$, so if we want the "normalized" form, we should add that extra factor.