For b) works Lemoine point

Let $d_a, d_b$ and $d_c$ be the distances from $P$ to the tangents to the circumcircle at $A$, $B$ and $C$ respectively, and let $\angle PAB = \alpha$, $\angle PBC = \beta$ and $\angle PCA = \gamma$. Finally, let $D,E,F$ be the foot from $P$ to sides $BC, CA, AB$. We have $$ d_a = AP \cdot \sin (C + \alpha),~~~~~~ d_b = BP \cdot \sin (A+\beta) ~~~~\text{and}~~~~ d_c = CP \cdot \sin (B+\gamma)$$ and also $$PE = AP \cdot \sin (A-\alpha) ~~~\text{and}~~~ PF = AP \cdot \sin (\alpha) \Rightarrow \sqrt{PE \cdot PF} = AP \cdot \sqrt{\sin (\alpha) \sin (A - \alpha)}$$ Multiplying cyclically the above expressions and using trig ceva, we want $$d_a d_b d_c \geq 8 \cdot PD \cdot PE \cdot PF $$ $$\iff \sin (C+ \alpha) \sin (A+ \beta) \sin (B+ \gamma) \geq 8 \sqrt{\sin (\alpha) \sin (A - \alpha)\sin (\beta) \sin (B - \beta)\sin (\gamma) \sin (C - \gamma)} = 8 \cdot \sin (\alpha) \sin (\beta) \sin (\gamma)$$ Now, let $Q$ be the isogonal conjugate of $P$ wrt $\triangle ABC$. Let $AQ$, $BQ$ and $CQ$ intersect sides $BC$, $CA$, and $AB$ in $X$, $Y$ and $Z$, respectively. Thus, by law of sines in $\triangle AXC$ and the analogous, we have $$\left( \dfrac{\sin (C+ \alpha)}{\sin (\alpha)}\cdot \dfrac{\sin (A+ \beta)}{\sin (\beta)} \cdot \dfrac{\sin (B+ \gamma)}{\sin (\gamma)} \right) ^2 = \left( \dfrac{AC}{CX} \cdot \dfrac{AB}{AY} \cdot \dfrac{BC}{BZ} \right) ^2 = \dfrac{BC^2}{CX(BC-CX)} \cdot \dfrac{AC^2}{AY(AC-AY)} \cdot \dfrac{AB^2}{BZ(AB-BZ)} \geq 4^3 = 64, $$ Where we're using ceva and $AM-GM$ at the end. Hence, taking the sqrt we're done. PS: Equality holds in AM-GM iff $CX = \frac{1}{2} BC$, $AY = \frac{1}{2} AC$ and $BZ = \frac{1}{2} AB$, i.e, $\iff Q $ is the centroid $\iff P$ is the lemoine point :0

We will use the method of Trilinear coordinates. Let the point $P$ have actual coordinates $d:e:f$. Then the tangent at $A$ has equation $cy+bz=0$. Using the point to line distance we compute that the distance from $P$ to the tangent at $A$ is given by $$\frac{ce+bf}{\sqrt{c^2+b^2-2cb\cos(A)}}=\frac{ce+bf}{a}$$Thus we need to show that $$8def\leq \left(\frac{ce+bf}{a}\right)\left(\frac{af+cd}{b}\right)\left(\frac{bd+ae}{c}\right)$$But this inequality is trivial by applying AM-GM to each of the sums. Notice that inequality holds when $d:e:f=a:b:c$, or at the symmedian point.