Easy right?

Lemma : If $x,y,z$ are real numbers and $xyz=1$ , we have $\displaystyle\sum_{cyc} \dfrac{1}{1+xy+x}=1$

Back to the original problem, let $a^{\frac{4}{3}}=x; b^{\frac{4}{3}}=y; c^{\frac{4}{3}}=z$ we get $xyz=1$ .

Applying AM-GM inequality, we have:

$ab+ab+ab+1\ge4\sqrt[4]{a^3b^3}\;\Rightarrow\;ab+\dfrac{1}{3}\ge\dfrac{4}{3}a^{\frac{4}{3}}b^{\frac{4}{3}}$

$a+a+a+1\ge4\sqrt[4]{a^3}\;\Rightarrow\;a+\dfrac{1}{3}\ge\dfrac{4}{3}a^{\frac{4}{3}}$

Thus, $\dfrac{1}{2+ab+a}\le\dfrac{1}{\dfrac{4}{3}+\dfrac{4}{3}a^{\frac{4}{3}}b^{\frac{4}{3}}+\dfrac{4}{3}a^{\frac{4}{3}}}=\dfrac{3}{4}.\dfrac{1}{1+xy+x}$

Similarly, we also get $\dfrac{1}{2+bc+b}\le\dfrac{3}{4}.\dfrac{1}{1+yz+y}$ and $\dfrac{1}{2+ca+c}\le\dfrac{3}{4}.\dfrac{1}{1+zx+x}$ .

Hence, $\displaystyle P=\sum_{cyc}\dfrac{1}{2+ab+a}\le\dfrac{3}{4}\sum_{cyc}\dfrac{1}{1+xy+x}=\dfrac{3}{4}$ .

So, $\max P=\dfrac{3}{4}=\boxed{0.75}$ if and only if $a=b=c=1$ .