First, we calculate cohomologies and find a representatives for even degree $n = 2p$.
Identify $U_{\alpha_0} \cong \mathbb C^m$ and let $z_1, \ldots, z_m$ be the standard coordinates. Observe that $U_{\alpha_0, \alpha_1, \ldots, \alpha_q} = \{z_{\alpha_i} \neq 0 | i>0\} \cong (\mathbb C\setminus \{0\})^q \cong (S^1)^q$ homotopically.
Suppose $\omega = (\omega^{0, n}, \ldots, \omega^{n, 0})$ be a $D$-cocycle in $K$.
Since $\delta$ is exact, we may choose a representative $\eta^{0, n}$ such that $[\omega] = [\eta^{0, n}]$.
On the other hand, we have $H^q(U_{\alpha_0\ldots\alpha_p}) \cong H^q((S^1)^p)$.
Also, $H^*{(S^1)^p} = H^*(S^1) \otimes \cdots \otimes H^*(S^1) = \otimes^{n}(\mathbb R \oplus \mathbb R)$ implies $H^q(U_{\alpha_0\ldots\alpha_p}) = 0$ for $q > p$.
Thus we can move $\eta^{0, n}$ to the diagonal of the double complex repeating the following process(as we take a representative $\omega^{0, n}$):
$D''\eta^{0, n} = 0$, and $H^n(U_{\alpha_0}) = 0 \implies \eta = D''\eta^{0, n-1}$ for some $\eta^{0, n-1}$...
Now, we have a representative $\eta^{p, p}$.
Choose a generator $d\theta_{ij} \in H^1(U_{i, j}) \cong H^1(S^1) \cong \mathbb R$ with relation $d\theta_{ij} + d\theta_{jk} + d\theta_{ki} = 0$ in the cohomology class of $H^1(U_{ijk})$.
Then $\wedge_{j=1}^{p} d\theta_{i_{\alpha_0}i_{\alpha_j}}$ is a generator for $H^{p}(U_{\alpha_0\ldots\alpha_p})$.
Thus, $\eta^{p, p} = \prod_{\alpha_0 \cdots \alpha_p} c_{\alpha_0\ldots\alpha_p} \wedge_{j=1}^p d\theta_{i_{\alpha_0}i_{\alpha_j}} + d\omega^{p, p-1}$ for some $\omega^{p, p-1}$.
We need $\delta\eta^{p, p}$ to be exact. i.e., $(\delta\eta^{p, p})_{\alpha_0\ldots\alpha_{p+1}} = \sum_{i=0}^{p+1}(-1)^i\eta^{p, p}_{\alpha_0\ldots\hat{\alpha_i}\ldots\alpha_{p+1}} = 0$ in $H^{p}(U_{\alpha_0\ldots\alpha_{p+1}}) \cong \mathbb R^p$.
Fix a trivialization $U_{\alpha_0} \cong \mathbb C^{m}$ and identify $U_{\alpha_0\ldots\alpha_{p+1}} = \{z_{\alpha_i} \neq 0 | i>0\}$ and $H^p(U_{\alpha_0\ldots\alpha_{p+1}}) \cong H^p((S^1)^{p+1}) \cong \mathbb R^p$.
Then $U_{\alpha_0\ldots\alpha_{p+1}} \to U_{\alpha_0\ldots\hat{\alpha_{i}}\ldots\alpha_{p+1}}$ induces the cohomology map $\mathbb R \to \mathbb R^p$ to the $i$th component.
Thus, the generators for $H^p(U_{\alpha_0\ldots\hat{\alpha_{i}}\ldots\alpha_{p+1}})$ form a basis for $H^p(U_{\alpha_0\ldots\alpha_{p+1}})$.
On the other hand, we have a nontrivial relation as
\begin{align*}
d\theta_{\alpha_1\alpha_2}\wedge d\theta_{\alpha_1\alpha_3} \wedge \cdots \wedge d\theta_{\alpha_1\alpha_{p+1}} & = (d\theta_{\alpha_0\alpha_2}-d\theta_{\alpha_0\alpha_1})\wedge \cdots \wedge (d\theta_{\alpha_0\alpha_{p+1}}-d\theta_{\alpha_0\alpha_1}) \\
& = \sum_{i=1}^{p+1} (-1)^{i+1} d\theta_{\alpha_0\alpha_1}\wedge \ldots \hat{d\theta_{\alpha_0\alpha_i}}\ldots d\theta_{\alpha_0\alpha_{p+1}} \\
\end{align*}
Hence, if we fix the coefficient $c_{0\ldots p}$, then the other coefficients are automatically determined by the above relation.
Furthermore, $\eta = \prod_{\alpha_0\ldots\alpha_p} \wedge_{j=1}^p d\theta_{i_{\alpha_0}i_{\alpha_j}}$ is a closed form by the above relation.
Thus there is only one cohomology class $[\eta]$ in $H^p(U_{\alpha_0\ldots\alpha_p})$, which implies $\dim H^{2p}(\mathbb C P^m) = 1$.
For the odd cohomology, let $0 \to C^0 \xrightarrow{D^0} C^1 \xrightarrow{D^1} C^2 \cdots$ be a $D$-cochain complex.
Generally, we have
\begin{align*}
\dim H^1 + \dim H^3 + \cdots & = \dim \ker D^1 - \dim \text{im} D^0 + \dim \ker D^3 - \dim \text{im} D^2 + \cdots \\
\dim H^0 + \dim H^2 + \cdots & = \dim \ker D^0 + \dim \ker D^2 - \dim \text{im} D^1 + \dim \ker D^3 - \cdots = m+1
\end{align*}
Subtract the first equation from the second, we have
\begin{align*}
\dim H^1 + \dim H^3 + \cdots = m+1 - \dim C^0 + \dim C^1 - \dim C^2 + \cdots
\end{align*}
The dimension of $C^n$ is given by
\begin{align*}
\dim C^n = \sum_{p+q = n} \binom{m+1}{p+1}\binom{p}{q}.
\end{align*}
Thus, $\dim H^1 + \dim H^3 + \cdots = m+1 - \dim C^0 + \dim C^1 - \dim C^2 + \cdots = 0$ and $\dim H^{2p+1}(\mathbb C P^m) = 0$.
