Let $X$ be a compact Kaehler manifold with $D$ be an effective divisor on $X$ such that $K_X+D$ is semiample and big then $X$ is projective?

3$\begingroup$ I believe that is not true. My recollection is that there exists a projective CalabiYau threefold $Y$ and a collection of smooth, disjoint rational curve $C_i\cong \mathbb{P}^1$ with normal bundle $\mathcal{O}(1)^{\oplus 2}$ such that the "flop" $X'$ of all of these curves is projective, yet the "flop" $X$ of some proper subset is a nonprojective algebraic space. Defining $D$ to be the pullback of an ample divisor from the contraction of all curves $C_i$ should give a counterexample. $\endgroup$ – Jason Starr May 27 '16 at 10:42

6$\begingroup$ If $X$ admits a big line bundle, then it is Moishezon. Moreover, a compact Moishezon manifold that admits a Kähler metric is projective. $\endgroup$ – Henri May 27 '16 at 13:15

$\begingroup$ @Henri: oops, I totally forgot about the Kaehler part. I guess I got too excited about working out my example! $\endgroup$ – potentially dense May 27 '16 at 13:19
Edit: I forgot about the Kaehler condition, so my answer is not relevant to the OP's question. As the comment of Henri shows, the correct answer is "yes". I will leave my original answer here (in modified form) in case anyone finds it useful.
The comment of Henri shows the answer is yes. One could ask: is the same true without the Kaehler condition? Let me give an example showing that is not true, based on Jason Starr's idea in the comment.
In $\mathbf P^3$ take 4 points $p_1, p_2, p_3, p_4$ in general position. Let $X$ be the blowup at these 4 points.
Denote by $L_{ij}$ the proper transform on $X$ of the line through $p_i$ and $p_j$. Note that these curves are all smooth and rational, have normal bundle $O(1) \oplus O(1)$, and are disjoint.
Next, there is a birational contraction morphism $X \rightarrow Z$ to a projective variety $Z$ that contracts all the $L_{ij}$. This is given by the linear system of quadrics through all the points $p_i$ (or more precisely, the proper transforms of those quadrics on $X$).
Now let $X'$ be the flop of the lines $L_{12}$ and $L_{34}$. (Note that the evident rational map $X' \dashrightarrow Z$ is in fact a morphism $X' \rightarrow Z$, because the indeterminacy locus of the flop is contracted to a pair of points on $Z$.)
Let $A$ be an ample divisor on $Z$, $\Delta$ its pullback to $X'$, and $D=K_{X'}+\Delta$. Then $D$ is effective and $K_{X'}+D=\Delta$ is semiample and big, as required.
I claim that $X'$ cannot be projective. The key point is that the numerical classes $[L_{ij}]$ satisfy the identities
$$[L_{ij}]+[L_{km}]=[L_{ik}]+[L_{jm}]$$ where $(i,j,k,m)$ is any permutation of $(1,2,3,4)$. This is easy to see: just write everything in the standard basis of $N_1(X)$.
Now the two "gluedin" curves on $X'$, which we denote by $\Lambda_{12}$ and $\Lambda_{34}$, have numerical classes $[\Lambda_{ij}]=[L_{ij}]$. On the other hand the flops do not change the numerical class of curves disjoint from the flopping curves, so the proper transforms of $L_{13}$ and $L_{24}$ on $X'$ still have classes $[L_{13}]$ and $[L_{24}]$. So the union of these four curves on $X'$ is an effective curve with numerical class
$$[\Lambda_{12}]+[\Lambda_{34}]+[L_{13}]+[L_{24}] = 0.$$
So $X'$ cannot have an ample divisor.

3$\begingroup$ I find it useful. If nothing else, this shows why the category of Kaehler manifolds is the wrong setting for doing birational geometry. $\endgroup$ – Jason Starr May 27 '16 at 13:27