Thanks for any help or comments

How can I find the list of all non abelian simple groups (particularly simple lie type) such that all $p$-Sylow subgroups are non abelian for odd prime $p$?

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Perhaps I will write an answer ( which I will mark as Community Wiki) to highlight some of the issues. From the Classification of Finite Simple Groups it is known that every finite simple group $G$ has at least one cyclic Sylow $p$-subgroup ( for an odd prime $p$ which really does divide $|G|$), which is a stronger assertion than required to answer the question. The Alternating Groups are easy to check as noted in comments, and the Sporadic Groups can be checked by easy inspection ( in the Atlas, for example).

Others are better qualified than I am to give a case by case check of the groups of Lie Type, but an analysis of the groups ${\rm PSL}(n,q)$ illustrates both why a positive answer might be expected and why it is not clear that a cyclic Sylow $p$-subgroup would necessarily be expected to have prime order ( and Derek Holt's examples of various Symplectic Groups indeed show that this need not be the case).

If we consider $H = {\rm GL}(n,q)$, then $H$ contains a Singer cycle of order $q^{n}-1,$ which gives, in particular, a cyclic subgroup of that order. By Zsigmondy's Theorem, except when $n =2$ or $n =6$ and $q = 2$, there is a prime $r$ which divides $q^{n}-1$ but does not divide $q^{i}-1$ for $ 1 \leq r < n.$ From this it follows ( apart from the exceptional cases) that $H$ has a cyclic Sylow $r$-subgroup and that ${\rm PSL}(n,q)$ has a non-trivial cyclic Sylow $r$-subgroup.

If $n =2$, then any simple ${\rm PSL}(2,q)$ has cyclic Sylow $r$-subgroups for any odd prime $r$ which divides $q^{2}-1,$ and there is at least one such prime $r$ as $q >3$ when ${\rm PSL}(2,q)$ is simple.

If $n = 6$ and $q = 2,$ then ${\rm GL}(6,2)$ has a cyclic non-trivial Sylow $31$-subgroup ( this is not as ad hoc as it might appear- in general when $n >2$ it can be checked that ${\rm GL}(n,q)$ always has a cyclic Sylow $r$-subgroup for some "Zsigmondy prime divisor" $r$ of one of $q^{n}-1$ or $q^{n-1}-1).$

Notice however, that ( at least without some deeper number theoretic analysis) this "Zsigmondy method" of producing cyclic Sylow $r$-subgroups does not a priori guarantee that $r$ will divide the group order to the first power. For example, ${\rm GL}(5,3)$ has a cyclic Sylow $11$-subgroup of order $11^{2}.$ Looking at ${\rm GL}(4,3)$ shows that ${\rm GL}(5,3)$ does have a cyclic Sylow $5$-subgroup, but in general, I see no obvious reason why all Zsigmondy prime divisors of both $q^{n}-1$ and $q^{n-1}-1$ should not occur to at least the second power.

Later edit: continuing this methodology, or by a direct argument, it is easy to see that for each prime $r$ which divides $|{\rm GL}(n,q)|,$ but not $|{\rm GL}(\lfloor \frac{n}{2} \rfloor, q) |,$ the Sylow $r$-subgroup of ${\rm GL}(n,q)$ is cyclic, and for large $n$, it would seem very improbable that each such Sylow subgroup had order greater than $r$.

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