Solution:

a. Let us denote by $O$ the circumcenter of $ABC$, and let us denote by $\alpha, \beta, \gamma$ the angles of $ABC$ at vertices $A, B, C$, respectively.

The fundamental point is that triangles $IAA_{1}$ and $AIO$ are congruent. To prove this, observe that $IA=AI$ since it is the same side, and $IA_{1}=AO$ by construction. It therefore suffices to prove that the angles included between these pairs of sides are equal.

The computation of these angles is delicate, since it depends on the configuration we are in, and more precisely on the position of $O$ relative to lines $AI$ and $BI$ (there are thus four cases to examine). Let us consider the case, shown in the figure on the left, in which $O$ lies on the same side of $C$ with respect to both lines (the other cases, one of which is shown in the figure on the right, are treated in a substantially analogous manner). In this configuration, to compute $\angle OAI$ it suffices to observe that $\angle IAB=\alpha / 2$ (because $AI$ is a bisector) and $\angle OAB=90^\circ-\gamma$ (because triangle $AOB$ is isosceles on base $AB$ and has vertex angle equal to $2\gamma$, since central angles are twice the corresponding inscribed angles). By subtraction we then obtain that
$$\angle OAI=\angle OAB-\angle IAB=90^\circ-\gamma-\frac{\alpha}{2}$$
For the computation of $\angle A_{1}IA$ recall that the sum of the interior angles of the (non-convex) quadrilateral $AIEC$ must be $360^\circ$, from which (here by $\angle AIE$ we denote the angle of the quadrilateral, which is greater than $180^\circ$)
$$\angle AIE=360^\circ-\angle CAI-\angle IEC-\angle ECA=360^\circ-\frac{\alpha}{2}-90^\circ-\gamma,$$
and therefore by subtraction
$$\angle A_{1}IA=\angle AIE-180^\circ=360^\circ-\frac{\alpha}{2}-90^\circ-\gamma-180^\circ=90^\circ-\frac{\alpha}{2}-\gamma,$$
which completes the proof of the congruence.

Reasoning in the same way on the other side, we discover that triangles $IBB_{1}$ and $BIO$ are congruent. At this point, from the two congruences we deduce that
$$\angle IA_{1}P+\angle IB_{1}P=\angle IA_{1}A+\angle IB_{1}B=\angle AOI+\angle BOI=\angle AOB=2\gamma .$$
Let us now consider quadrilateral $A_{1}PB_{1}I$, and observe that $\angle A_{1}IB_{1}=\angle DIE=180^\circ-\gamma$ (by the cyclicity of $CDIE$, which has two opposite right angles). By subtraction we then obtain that
$$\begin{aligned} \angle APB & =\angle A_{1}PB_{1} \\ & =360^\circ-\angle A_{1}IB_{1}-\angle IA_{1}P-\angle IB_{1}P \\ & =360^\circ-\left(180^\circ-\gamma\right)-2\gamma \\ & =180^\circ-\gamma, \end{aligned}$$
and this shows that, in this configuration, the point $P$ lies on the circumscribed circle of triangle $ABC$, on the opposite side of line $AB$ from $C$.

In the configuration on the right the proof is analogous, but the computations are slightly different. In this case we will have that
$$\angle OAI=\angle IAB-\angle OAB=\frac{\alpha}{2}-90^\circ+\gamma$$
and (denoting this time by $\angle AIE$ the angle smaller than $180^\circ$)
$$\angle A_{1}IA=180^\circ-\angle AIE=180^\circ-\left(360^\circ-\frac{\alpha}{2}-90^\circ-\gamma\right)=\frac{\alpha}{2}+\gamma-90^\circ,$$
from which once again the congruence of triangles $IAA_{1}$ and $AIO$.

From the congruence we deduce again that $\angle IA_{1}P=\angle IA_{1}A=\angle AOI$. Unlike before, reasoning on the other side we now obtain that
$$\angle IB_{1}P=180^\circ-\angle IB_{1}B=180^\circ-\angle BOI,$$
from which
$$\angle IA_{1}P+\angle IB_{1}P=180^\circ-(\angle BOI-\angle AOI)=180^\circ-\angle AOB=180^\circ-2\gamma .$$
Let us now consider again the (this time non-convex) quadrilateral $A_{1}IB_{1}P$. As before we have that $\angle A_{1}IB_{1}=\angle DIE=180^\circ-\gamma$, and therefore $\angle A_{1}IB_{1}=180^\circ+\gamma$ as a non-convex angle. By subtraction we then obtain that
$$\begin{aligned} \angle APB & =\angle A_{1}PB_{1} \\ & =360^\circ-\angle IA_{1}P-\angle IB_{1}P-\angle A_{1}IB_{1} \\ & =360^\circ-\left(180^\circ-2\gamma\right)-\left(180^\circ+\gamma\right) \\ & =\gamma, \end{aligned}$$
and this shows that, in this configuration, the point $P$ lies on the circumscribed circle of triangle $ABC$, on the same side of line $AB$ as $C$.

b. The possible values of the perimeter of $ABC$ are all those (strictly) between $2$ and $3$.

To prove this, let us first observe that $P$ can coincide with $C$ only when we are in a configuration like the one in the figure on the right. Moreover, the points $A_{1}, A, C$ must be collinear, which tells us that $\angle A_{1}AI=180^\circ-\alpha / 2$, and consequently (because of the usual congruence) also $\angle AIO=180^\circ-\alpha / 2$. This implies that line $IO$ is parallel to line $AB$, with $I$ to the left of $O$. There also exists an analogous configuration in which $A_{1}$ lies between $A$ and $C$ and the equality $\angle A_{1}AI=\angle AIO=\alpha / 2$ holds, which tells us that line $IO$ is parallel to line $AB$ with $I$ to the right of $O$.

In any case, $I$ and $O$ are at the same distance from line $AB$. Now the distance of $I$ equals $r$, that is, the radius of the inscribed circle, while the distance of $O$ equals $R \cos \gamma$ (this can be seen by observing that the isosceles triangle $ABO$ can be decomposed into the union of two right triangles with angles at $O$ equal to $\gamma$ and hypotenuse of length $R$). The following equality must therefore hold
$$r=R \cos \gamma$$
Denoting by $S$ the area and by $p$ the semiperimeter, and recalling the well-known formulas
$$r=\frac{S}{p}, \quad \quad \quad \quad \cos \gamma=\frac{ab c}{4S}, \quad \frac{a^{2}+b^{2}-c^{2}}{2ab}$$
we can express the previous relation as
$$8S^{2}=p\left(a^{2}+b^{2}-c^{2}\right) .$$
If we now compute $S$ using Heron's formula, and simplify a factor of $p$, we finally obtain that
$$(a+b-c)(b+c-a)(c+a-b)=c\left(a^{2}+b^{2}-c^{2}\right),$$
which in the case $c=1$ reduces to
$$(a+b-1)(b+1-a)(1+a-b)=a^{2}+b^{2}-1 .$$
Setting $s=a+b$ and $d=a-b$, we can rewrite the equality in the form
$$\frac{s^{2}+d^{2}}{2}-1=(s-1)(1-d)(1+d)=(s-1)\left(1-d^{2}\right)$$
from which with simple algebraic steps we obtain that
$$d^{2}=\frac{s(2-s)}{2s-1} .$$
Since $d^{2} \geq 0$, from this relation it is evident that we must have $s<2$ (the case $s=2$ must also be excluded because it would lead to $d=0$, that is, to an equilateral triangle). On the other hand it is evident that we must also have $s>1$, because in every triangle the sum of the lengths of two sides is greater than the length of the third.

It remains to verify that for every $1<s<2$ we obtain an admissible triangle. In this case the lengths of the two sides $BC$ and $CA$ are
$$a=\frac{s+d}{2}, \quad b=\frac{s-d}{2},$$
or vice versa. Are these the sides of a triangle? It suffices that each side be smaller than the sum of the other two. We already know that $a+b>1$, and it is obvious that $a+1>b$. It remains therefore only to verify that $b+1>a$, but this inequality reduces to $d<1$, from which with simple steps we again find the condition $s>1$.