My proof consists in some very nasty algebraic manipulations, resulting from the indiscriminate use of barycentric coordinates; you have been warned. Let thus ${\rm P} = (u, v, w)$ be the normalized barycentric coordinates of the point $\rm P$ (that is, $u+v+w=1$), and recall that those of the remaining ones are $$ {\rm A} = (1, 0, 0),\qquad {\rm B} = (0, 1, 0),\qquad {\rm C} = (0, 0, 1),\qquad {\rm I} = \left(\frac a{a+b+c}, \frac b{a+b+c}, \frac c{a+b+c}\right). \tag1 $$ The distance formula for points $\rm P,Q$ in normalized barycentric coordinates requires taking the so-called displacement vector $\overrightarrow{\rm PQ} = (r,s,t)$ whose entries are the subtraction of those of $\rm P$ from those of $\rm Q$; then, it holds that $$ {\rm PQ}^2 = -a^2st - b^2tr - c^2rs. $$ Hence $\overrightarrow{\rm AP}=(u - 1, v, w), \overrightarrow{\rm BP}=(u, v - 1, w), \overrightarrow{\rm CP}=(u, v, w - 1)$, and expanding the left-hand side of the proposed identity yields $$ \begin{alignat*}2 a\cdot{\rm PA}^2 + b\cdot{\rm PB}^2 + c\cdot{\rm PC}^2 &= &&-a\left[a^2vw + b^2(u-1)w + c^2(u-1)v\right]\\ & &&- b\left[a^2(v-1)w + b^2 uw + c^2 u(v-1)\right]\\ & &&- c\left[a^2v(w-1) + b^2 u(w-1) + c^2 uv\right]\\ &= &&-a^3vw - ab^2uw + ab^2w - ac^2uv + ac^2v\\ & &&-a^2bvw + a^2bw - b^3 uw - bc^2uv + bc^2u\\ & &&-a^2cvw + a^2cv - b^2cuw + b^2cu - c^3uv; \tag2 \end{alignat*} $$ substituting the values of the last term in equation $(1)$ into the previous equation produces the first three terms of the right-hand side: $$ \begin{alignat*}2 a\cdot{\rm IA}^2 + b\cdot{\rm IB}^2 + c\cdot{\rm IC}^2 &= &&-a\cdot \frac{a^2bc - b^2c(b+c) - c^2b(b+c)}{(a+b+c)^2}\\ & &&- b\cdot \frac{ab^2c - a^2c(a+c) - c^2a(a+c)}{(a+b+c)^2}\\ & &&- c\cdot \frac{abc^2 - a^2b(a+b) - b^2a(a+b)}{(a+b+c)^2}\\ &= &&\frac{-1}{(a+b+c)^2}(\cancel{a^3bc} - \cancel{ab^3c} - ab^2c^2 - ab^2c^2 - \cancel{abc^3}\\ & &&\phantom{\frac{-1}{(a+b+c)^2}}+ \cancel{ab^3c} - \cancel{a^3bc} - a^2bc^2 - a^2bc^2 - abc^3\\ & &&\phantom{\frac{-1}{(a+b+c)^2}}+ \cancel{abc^3} - a^3bc - a^2b^2c - a^2b^2c - ab^3c)\\ &= &&\frac{abc\cancel{(a+b+c)^2}}{\cancel{(a+b+c)^2}} = {\color{green} \boxed{abc,}} \tag3 \end{alignat*} $$ and since $\overrightarrow{\rm IP} = (u-a/(a+b+c), v-b/(a+b+c), w-c/(a+b+c))$ it follows that $$ \begin{alignat*}2 (a+b+c){\rm IP}^2 &= &&-a^2\left[(a+b+c)vw - bw - cv + \frac{bc}{a+b+c}\right]\\ & &&-b^2\left[(a+b+c)uw - aw - cu + \frac{ac}{a+b+c}\right]\\ & &&-c^2\left[(a+b+c)uv - av - bu + \frac{ab}{a+b+c}\right]\\ &= &&-a^3vw - a^2bvw - a^2cvw + a^2bw + a^2cv - \frac{a^2bc}{a+b+c}\\ & &&-ab^2uw - b^3uw - b^2cuw + ab^2w + b^2cu - \frac{ab^2c}{a+b+c}\\ & &&-ac^2uv - bc^2uv - c^3uv + ac^2v + bc^2u - \frac{abc^2}{a+b+c}. \tag4 \end{alignat*} $$ Finally, notice that the left-hand side is contained in the $u,v,w$ terms of equation $(4)$, so it suffices to show that the remaining summands are cancelled by $(3)$. Indeed $$ abc - \frac{a^2bc + ab^2c + abc^2}{a+b+c} = abc - \frac{abc\cancel{(a+b+c)}}{\cancel{a+b+c}} = {\color{green} \boxed{0,}} $$ as it was to be proven.