The following proof shows that this is indeed the case. Let $\gamma$ denote the spectral gap of $P$ and $\lambda$ denote the absolute value of the eigenvalue of $P$ that achieves the spectral gap. Then,
\begin{equation}
\rho_1 \le 1 - \frac{\gamma}{2n}
\end{equation}
for large $n$.

*Proof:*

Let $\{ \mathbf{u}_i \}$ and $\{ \mathbf{v}_i \}$ respectively denote the set of eigenvectors of $P$ and $P'$ (both are symmetric matrices). Note that $\mathbf{u}_1 = \frac{1}{\sqrt{n}} \mathbf{1}$. In addition, let $\{ \lambda_1 (=1), \lambda_2, \cdots, \lambda_n \}$ be the set of eigenvalues of $P$ and $\{ \rho_1,\rho_2,\cdots,\rho_n\}$ denote the eigenvalues of $P'$.

Observe that,
\begin{align}
\mathbf{u}_1^T P' \mathbf{u}_1 &= \frac1n \left( \sum_{i,j \in [n]} P_{i,j} - \sum_{j \in [n]} P_{1,j} - \sum_{i \in [n]} P_{i,1} + P_{1,1} \right) \\
&= \frac1n \left( n - 2 + P_{1,1} \right) \\
&\le 1 - \frac1n \tag{1} \label{eq:1}
\end{align}

By the eigenvalue decomposition of $P'$,
\begin{align*}
\mathbf{u}_1^T P' \mathbf{u}_1 = \sum_{i \in [n]} \rho_i \langle \mathbf{u}_1, \mathbf{v}_i \rangle^2 \ge \rho_1 \langle \mathbf{u}_1, \mathbf{v}_1 \rangle^2 - \lambda (1 - \langle \mathbf{u}_1, \mathbf{v}_1 \rangle^2)
\end{align*}
note that Cauchy interlace theorem implies that $\forall i \ge 2,\ |\rho_i| \le \lambda$.

Combining with eq. \eqref{eq:1}, we have that,
\begin{equation}
(\rho_1 + \lambda) \langle \mathbf{u}_1, \mathbf{v}_1 \rangle^2 \le 1 + \lambda - \frac1n \tag{2} \label{eq:2}
\end{equation}

Now observe that by the Perron Frobenius theorem $\mathbf{v}_1$ can be assumed to have all positive entries. Therefore,
\begin{align}
\rho_1 = \mathbf{v}_1^T P' \mathbf{v}_1 \le \mathbf{v}_1^T P \mathbf{v}_1 &= \sum_{i \in [n]} \lambda_i \langle \mathbf{u}_i, \mathbf{v}_1 \rangle^2 \\
&\le \langle \mathbf{u}_1, \mathbf{v}_1 \rangle^2 + \lambda (1 - \langle \mathbf{u}_1, \mathbf{v}_1 \rangle^2) \\
\implies \rho_1 - \lambda &\le (1 - \lambda) \langle \mathbf{u}_1, \mathbf{v}_1 \rangle^2 \tag{3} \label{eq:3}
\end{align}

Multiplying eqs. \eqref{eq:2} and \eqref{eq:3},
\begin{equation}
\rho_1^2 - \lambda^2 \le (1 - \lambda) \left( 1 + \lambda - \frac1n\right)
\end{equation}
Therefore,
\begin{align}
\rho_1^2 &\le 1 - \frac{1-\lambda}{n} \\
\implies \rho_1 &\overset{(i)}{\le} 1 - \frac{1 - \lambda}{2n}
\end{align}
where $(i)$ holds for large $n$.