Using Biot Savart or Ampere's Law you will come to the same problem $B$ is not
defined on the ring.
This is the same problem that trying to find the Electric field $E$ of a puntual charge just in the point where the charge is placed $1/r²$ becomes $\infty$...
You need to use the formula for volumes but using the superficial current density $J$ and integrating on a torus, then the magnetic field is well defined. Notice that:
$\frac{\mu_0}{4\pi}\int_V \frac{Jdv\times r}{|r|^3}=\frac{\mu_0}{4\pi}\int_V \frac{4\pi r² d\Omega dr J\times r}{|r|^3}=\mu_0\int_V \frac{J\times u_r r³ d\Omega dr }{|r|^3}=\mu_0\int_V J\times u_r d\Omega dr$
Thus even if $r \to 0$ the $\infty$ does not appear.
The problem is that solving volume integrals is more complicated that using a line... but in this case I cannot find a better option.
I really like the proof contained in the paper Derivation of the Biot-Savart Law from Ampere's Law Using the Displacement Current from Robert Buschauer (2013)
It's simple and it fulfills the role of convincing the reader.
Basically the author works with one point charge $q$ situated in origin of Z azis $(0,0,0)$. He supposes a particle moving in Z axis to positive Z values with velocity $v$. He creates a magnetic field line in a arbitrarious circle with $c$ radius, by symmetry, with center in $(0,0,a)$. The angle between any point in the circle and the center of circle starting from origin $(0,0,0)$ is $\alpha$.
Starting point is a part of 4th Maxwell's Equation of electromagnetism, the Ampere-Maxwell Law that consider changing electric flow with time in a area produces magnetic field circulation. This law generates a magnetic force that can be verified using special relativity that in another reference frame it's just a plain electric force.
$$\oint B\, dl = \mu_0\epsilon_0 \; d/dt(\int_A E.dA)$$
In the left side, the solution consists of integrating the $\oint B dl$ in this circle (butterfly net ring). As $B$ is constant by symmetry, we have
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \oint B\, dl = 2\pi c B \qquad\qquad$ (1)
In the right side $\;[\;\mu_0 \epsilon_0 d/dt (\int_A E\; dA \,)\;],\;$ as the surface (butterfly net) we choose a sphere of radius $r$, to ensure that all points have the same value of electric field:
$$ E = q / 4\pi\epsilon_0r^2$$
Let's first calculate the right-hand integral in the right side. We adopted here a slightly different standard in spherical coordinate. Just to remember,the element for integration into spherical coordinates is $\; r^2 \sin \phi \, dr \, d\phi \, dq $
Let $\theta$ (XY axis) vary from $0$ to $2\pi$ and by consider the angle $\phi$ with the vertical (Z axis) from 0 to $\alpha$.
$$\Phi_E = \int_A E\; dA = q/4\pi \epsilon_0 r^2 \int_A dA = q/(4\pi \epsilon_0 r^2) r^2 \int_{0,2\pi} d\theta \int_{0,\alpha} \sin \Phi\; d\Phi = $$
$$q/4\pi \epsilon_0 2\pi ( -\cos \alpha + 1) = q/2\epsilon_0 (1 - cos\alpha)$$
Thus
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Phi_E = \mu_0 q /2 (1 - cos\alpha)$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad d \Phi_E / dt = - q/2\epsilon_0 d \cos \alpha/dt\qquad$(2)
Putting $\alpha$ as a function of $z$, we have, by the chain rule:
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad d \cos \alpha/dt = (d \cos\alpha/dz) \; (dz/dt)\qquad$(3)
However as $z$ is decreasing with the motion at velocity $v$, we have
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad dz / dt = -v \qquad $(4)
On the other hand:
$$ \cos \alpha = z / r = z / \sqrt{c^2 + z^2}$$
Using this online tool for derivation:
$d \cos \alpha/dz = c^2/r^3$ where $r = \sqrt{c^2 + z^2}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad 2\pi c B = q \mu_0 /2 v (c^2/r^3)\qquad$ By (1),(2),(3),(4)
$$B = \mu_0 q v c / 4\pi c r^3$$
but $\quad\sin \alpha = c / r\quad$ so we can add $\quad \sin \alpha\; r / c$:
$$B = \mu_0 q v \sin \alpha /4\pi r^2 $$
Vectorizing we have a cross product:
$$B = \mu_0 q \; v\uparrow \times r\uparrow /4\pi r^3$$
In some infinitesimal point we can consider a element of electric current as a point charge, so we can add other charge points by integration (any force is addictive!) for using in real applications. Thus we have in scalar notation:
$$dB = \mu_0 dq \; v \; r \sin \alpha /4\pi r^2$$
Considering $\quad dq = i\;dt\quad$ and $\quad v = ds/dt\quad $, we finally have reached to Biot-Savart law:
$$dB = \mu_0 i \; ds \; r \sin \alpha /4\pi r^2$$
Best Answer
One can make an educated guess that the magnetic field at $P$ is one-half the value of the magnetic field at the center of a current loop which is given by:
$$B = \frac{\mu_0 I}{2R} $$
So your first approach is correct.
The approach of your solution 2 doesn't make any sense to me. The integral is along a closed path and, evidently, you're assuming $\mathbf B \cdot d\mathbf l$ is a constant along a closed path of constant $r$?