I'll focus on the two steps that you asked for clarification on. For reference, the matrix being triangularized is
$$
A =
\begin{pmatrix}
3 & 0 & 1 \\
-1 & 4 & -3 \\
-1 & 0 & 5 \\
\end{pmatrix}.
$$
For the step of selecting a $\lambda$, we are trying to find an elementary matrix
$$
E_{\lambda} = \pmatrix{1 & 0 & 0\\0 & 1 & 0 \\
\lambda & 0 & 1}
$$
such that $E_\lambda AE_{\lambda}^{-1}$ has zero as its bottom-left entry. To that end, we compute the lower-left entry of this product. Taking $e_1,e_2,e_3 \in \Bbb R^3$ to be the standard basis (i.e. the columns of the identity matrix $I_3$), we have
$$
[E_\lambda AE_\lambda^{-1}]_{3,1} = \\
e_3^T[E_\lambda AE_\lambda^{-1}]e_1 = \\
(e_3^TE_\lambda) A(E_\lambda^{-1}e_1) = \\
\pmatrix{\lambda & 0 & 1}A \pmatrix{1\\0\\-\lambda} = \\
\pmatrix{\lambda & 0 & 1}
\pmatrix{3 - \lambda\\
-1 + 3\lambda\\
-1 - 5\lambda} = \\
\lambda(3-\lambda) - 1 - 5\lambda =\\
-\lambda^2 - 2\lambda - 1.
$$
As the answer states, in order to have this result equal zero, we solve $-\lambda^2 - 2\lambda - 1 = 0$ to get $\lambda = -1$. With that done, we now compute
$$
E_{\lambda}AE_{\lambda}^{-1} = \pmatrix{4 & 0 & 1 \\
-4 & 4 & -3 \\
0 & 0 & 4 \\}.
$$
Now, let $E_2$ denote the elementary matrix that swaps the first two rows. We have
$$
E_2 = E_2^{-1} = \pmatrix{0&1&0\\1&0&0\\0&0&1}.
$$
Applying the associated paired operation to this new matrix gives us something triangular.
$$
E_2[E_{\lambda}AE_{\lambda}^{-1}]E_2^{-1} =
\pmatrix{0&1&0\\1&0&0\\0&0&1}
\pmatrix{4 & 0 & 1 \\
-4 & 4 & -3 \\
0 & 0 & 4 \\}
\pmatrix{0&1&0\\1&0&0\\0&0&1} \\
= \pmatrix{4 & -4 & -3 \\
0 & 4 & 1 \\
0 & 0 & 4 \\}.
$$
Now, note that this product can be written as
$$
\pmatrix{4 & -4 & -3 \\
0 & 4 & 1 \\
0 & 0 & 4 \\} = (E_2 E_{\lambda}) A (E_2 E_{\lambda})^{-1}.
$$
That is, if we take $P = (E_2 E_\lambda)^{-1} = E_{\lambda}^{-1}E_2^{-1}$, then the upper triangular matrix that we obtained can be written as $P^{-1}AP$, which is what we wanted.
