The question is essentially about how to relate the unitary/isometric representation of the dephasing channel with its representation in terms of Kraus operators.
In general, an isometric representation of a channel $\Phi$ is a writing of the form $\Phi(\rho)=\operatorname{tr}_2[V\rho V^\dagger]$ for some isometry $V$. This is the representation you're starting with: you're describing/defining the amplitude damping channel as the isometry defined in the computational basis as
$$V|0\rangle= |0\rangle\otimes|0\rangle,
\qquad
V|1\rangle = \sqrt{1-p}|1\rangle\otimes|0\rangle+\sqrt p |0\rangle\otimes|1\rangle.$$
Note that this is an isometry because it's a linear operator of the form $V:\mathbb{C}^2\to(\mathbb{C}^2)^{\otimes2}$ such that $V^\dagger V=I_2$. Equivalently, it's a matrix with orthonormal columns.
As a matrix you can represent this as
$$V = \begin{pmatrix}
1 & 0 \\
0 & \sqrt p \\ 0 & \sqrt{1-p} \\ 0&0
\end{pmatrix}.$$
Given any such isometric representation, you can find a Kraus representation for $\Phi$ using the operators $A_a\equiv (I\otimes \langle a|)V$, with $|a\rangle$ a(ny) orthonormal basis for the environmental space (the second space used in the definition of $V$). See also this and this for more details about the general proceedure.
In the case at hand, this gives
$$A_0 = (I\otimes \langle0|)V = \begin{pmatrix}1&0\\0&\sqrt{1-p}\end{pmatrix},
\qquad
A_1 = (I\otimes \langle1|)V = \begin{pmatrix}0&\sqrt p\\0&0\end{pmatrix}.$$
Finding a different set of Kraus operators
For the sake of completeness, note that you can also find different Kraus operators from this isometric representation, using different bases in the formula above. For example, using the eigenbasis of the Pauli $X$ matrix, you get
$$A_0' \equiv (I\otimes \langle +|)V
= \frac1{\sqrt2}\begin{pmatrix}1 & \sqrt p \\ 0 & \sqrt{1-p}\end{pmatrix},
\qquad
A_1' \equiv (I\otimes\langle -|) V
= \frac{1}{\sqrt2}\begin{pmatrix}1 & -\sqrt p\\ 0 & \sqrt{1-p}\end{pmatrix}.$$
Note that these are related to the previous Kraus operators via
$\sqrt2 A_0'=A_0 + A_1$ and $\sqrt2 A_1'=A_0-A_1$.
Prove equivalence of Kraus decompositions
Though they look different, these Kraus operators describe the same identical channel as the ones derived above. And the same applies to any set of Kraus operators derived this way.
You can, for example, verify this by showing that both Kraus decompositions correspond to the same Choi matrix, which you obtain by computing $J(\Phi)=\sum_a \operatorname{vec}(A_a)\operatorname{vec}(A_a)^\dagger$ with $\operatorname{vec}(A_a)$ vectorisation of $A_a$. For more details on this process see e.g. this and this. You'll find that regardless of which Kraus decomposition you use you'll get
$$J(\Phi)
= \begin{pmatrix}1 & 0 & 0 &\sqrt{1-p} \\ 0&p&0&0\\0&0&0&0\\\sqrt{1-p}&0&0&1-p\end{pmatrix}.$$
It's easy to verify that this Choi has eigendecomposition
$$J(\Phi) = (2-p) \mathbb{P}_u + p \mathbb{P}_v,
\qquad \mathbb{P}_u\equiv |u\rangle\!\langle u|, \\
|u\rangle \equiv \frac{1}{\sqrt{2-p}}\begin{pmatrix}1\\0\\0\\\sqrt{1-p}\end{pmatrix},
\qquad |v\rangle\equiv\begin{pmatrix}0\\1\\0\\0\end{pmatrix}.$$
Remember that rank-1 decompositions of the Choi corresponding bijectively to Kraus decompositions of the channel.
The first Kraus decompositions above corresponds to the eigendecomposition of the Choi. It is however interesting to note that the second decomposition does not. It instead corresponds to a different rank-1 decomposition, namely,
$$J(\Phi) =
\mathbb{P}_{w_1} + \mathbb{P}_{w_2}, \\
|w_1\rangle \equiv \frac{1}{\sqrt2}(\sqrt{2-p} |u\rangle + \sqrt p |v\rangle),
\qquad
|w_2\rangle \equiv \frac{1}{\sqrt2}(\sqrt{2-p} |u\rangle - \sqrt p |v\rangle).$$
You can verify directly that if you unvectorise these two (unnormalised) vectors you get precisely $A_0',A_1'$.
You can more characterise all possible rank-1 decompositions of a positive semidefinite operators (and thus all possible Kraus decompositions of a channel) as discussed e.g. here and here.
