 Solution

The exceptional divisor is $E\cong {\mathbb{P}}^{r-1}$, so its Chow groups are generated by the linear subspaces $[{\Gamma }_k]$. The complement $X\setminus E$ is isomorphic through the blow-up map to ${\mathbb{P}}^r\setminus \{p\}$. The localization sequence for $\{p\}\subseteq {\mathbb{P}}^r$ shows that the latter's Chow groups are generated by the restrictions of linear spaces $H_j$ through $p$ for $1\le j\le r$; its zero-dimensional Chow group vanishes.

Under the complement isomorphism, $H_j\setminus \{p\}$ corresponds to the open part of the strict transform ${\tilde{H}}_j$. A second localization sequence, now for $E\subseteq X$, shows that $A_{\ast }(X)$ is generated by the lifts $[{\tilde{H}}_j]$ together with the images $[{\Gamma }_k]$ from $A_{\ast }(E)$, exactly as claimed.

Solved by gpt-5.6-sol high.