Let $(M,\omega)$ be a compact symplectic manifold and the cohomology class $$[\omega]+\frac{1} {2}c_1(\wedge_{\mathbb C}^{0,n}(TM, J))\in H_{dR}^2(M)$$ is integral, for some almost complex structure $J$ on $M$ .Then why there exists a $Spin^c$-structure $P\to M$ on $M$? AND why a $\text{spin}^c$ structure with determinant $L^{2ω}$ exists on $(M,ω)$. I saw it in the thesis of Peter Hochs In Page 43.

You ask "why does a spin$^c$ structure exist on $(M,\omega)$?" It is a basic fact that every symplectic manifold admits a spin$^c$ structure. I will assume that you are instead interested in what Hochs is claiming in his comment on page 43 of his thesis: a spin$^c$ structure with determinant $L^{2\omega}$ exists on $(M,\omega)$.

The answer is related to an answer I gave to a previous question of yours. Let me quote the relevant part:

Complex line bundles over a manifold are classified by their first Chern class; we have a bijection $$\{\text{isomorphism classes of complex line bundles on $X$}\} \leftrightarrow H^2(X; \Bbb Z),$$ $$L \mapsto c_1(L).$$ The first Chern class is additive with respect to tensor products, so we see that $$c_1(L^{\otimes 2}) = 2c_1(L). \tag{$\ast$}$$

On page 43 Hochs's thesis, it is claimed that under the integrality condition posed in your question, the line bundle $$L^{2\omega} \otimes \wedge_{\Bbb C}^{0, \dim(M)} (TM, J) \longrightarrow M$$ has a square root. Here $L^{2\omega}$ is a line bundle with first Chern class $2[\omega]$.

By my quoted answer, since $[\omega]+ \tfrac{1}{2}c_1(\wedge_{\Bbb C}^{0, \dim(M)} (TM, J))$ is integral, there exists a complex line bundle $\mathcal{L} \longrightarrow M$ such that $$c_1(\mathcal{L}) = [\omega]+ \tfrac{1}{2}c_1(\wedge_{\Bbb C}^{0, \dim(M)} (TM, J)).$$ Furthermore, we have that \begin{align*} c_1(\mathcal{L}^{\otimes 2}) & = 2[\omega]+ c_1(\wedge_{\Bbb C}^{0, \dim(M)} (TM, J)) \\ & = c_1(L^{2\omega} \otimes \wedge_{\Bbb C}^{0, \dim(M)} (TM, J)). \end{align*} Hence $\mathcal{L}$ is a square root of $L^{2\omega} \otimes \wedge_{\Bbb C}^{0, \dim(M)} (TM, J)$ (up to minor annoyances due to $2$-torsion in $H^2(M;\Bbb Z)$, which can be dealt with).

Now, if you look at Proposition D.50 in the reference Hochs mentions, it tells us that on the complex tangent bundle $(TM, J)$, we get a spin$^c$ structure just by choosing a line bundle on $M$. Take $\mathcal{L}$ as described above to be our complex line bundle, and let $\sigma(\mathcal{L})$ denote the associated spin$^c$ structure. Proposition D.50 further says that $$\det(\sigma(\mathcal{L})) \cong K^\ast \otimes \mathcal{L}^{\otimes 2},$$ where $K^\ast$ is the anticanonical line bundle of $(M,J)$. Since $\mathcal{L}^{\otimes 2} \cong L^{2\omega} \otimes \wedge_{\Bbb C}^{0, \dim(M)} (TM, J)$ and $K^\ast \otimes \wedge_{\Bbb C}^{0, \dim(M)} (TM, J)$ is the trivial line bundle, we find that $$\det(\sigma(\mathcal{L})) \cong L^{2\omega}.$$ The remark in Hochs's thesis you are asking about is just saying that this spin$^c$ structure on $M$ with determinant $L^{2\omega}$ exists.