Let:
$A$ represent $\ce{PCl5}$
$C$ represent $\ce{PCl3}$
$D$ represent $\ce{Cl2}$
Then the reaction becomes:
$$\ce{A <=> C +D}$$
Let:
(1) represent "before removing $\ce{Cl2}$"
(2) represent "after removing $\ce{Cl2}$"
Assuming that, initially, we have $P_{A0}$ amount of $A$ and no $C$ or $D$ present:
$$P{A_1= P_{A0}-x}$$
$$P{C_1=0+x}$$
$$P{D_1=0+x}$$
$$---------$$
$$P_1=P_{A0}+x$$
Then, we remove all the $\ce{Cl2}$, so more products are formed:
$$P{A_2=P_{A0}-x-y}$$
$$P{C_2=x+y}$$
$$P{D_2=y}$$
$$---------$$
$$P_2=P_{A0}+y$$
Then, we define the difference in total equilibrium pressure as:
$$\Delta P=P_2-P_1=y-x$$
Temperature is constant, so $K_P$ won't change after removing $\ce{Cl2}$:
$$K_P=\frac{P_{C2}\;P_{D2}}{P_{A2}}=\frac{P_{C1}\;P_{D1}}{P_{A1}}$$
Rearranging:
$$\left(\frac{P_{C2}}{P_{C1}}\right)\left(\frac{P_{D2}}{P_{D1}}\right)=\left(\frac{P_{A2}}{P_{A1}}\right)$$
Substituting in terms of $x$, $y$, and $P_{A0}$:
$$\left(\frac{x+y}{x}\right)\left(\frac{y}{x}\right)=\left(\frac{P_{A0}-x-y}{P_{A0}-x}\right)$$
Now, we define the dissociation fractions of $A$ for equilibrium system (1) and (2), respectively as:
$$\alpha=\frac{x}{P_{A0}}$$
$$\beta=\frac{y}{P_{A1}}=\frac{y}{P_{A0}-x}$$
Solving for $x$ and $y$:
$$x=P_{A0}\;\alpha$$
$$y=P_{A0}\;\beta-x\;\beta=P_{A0}\;\beta-P_{A0}\;\alpha\;\beta$$
Substituting $x$ and $y$ in the expression we derived earlier:
$$\left(\frac{P_{A0}\;\alpha+P_{A0}\;\beta-P_{A0}\;\alpha\;\beta}{P_{A0}\;\alpha}\right)\left(\frac{P_{A0}\;\beta-P_{A0}\;\alpha\;\beta}{P_{A0}\;\alpha}\right)=\left(\frac{P_{A0}-P_{A0}\;\alpha-P_{A0}\;\beta-P_{A0}\;\alpha\;\beta}{P_{A0}-P_{A0}\;\alpha}\right)$$
All the $P_{A0}$ terms cancel out:
$$\left(\frac{\alpha+\beta-\alpha\;\beta}{\alpha}\right)\left(\frac{\beta-\alpha\;\beta}{\alpha}\right)=\left(\frac{1-\alpha-\beta-\alpha\;\beta}{1-\alpha}\right)$$
After expanding, we get:
$$\alpha^3-\alpha^3\beta^2+3\alpha^2\beta^2-\alpha^2\beta-\alpha^2-3\alpha\beta^2+\alpha\beta+\beta^2=0$$
The positive solution for $\beta$ in terms of $\alpha$ is:
$$\beta=\frac{\alpha\left(\sqrt{4\alpha^2-8\alpha+5}-1\right)}{2(1-\alpha)^2}=f(\alpha)$$
Recalling our derived expression for $\Delta P$:
$$\Delta P=y-x=P_{A0}\;\beta-P_{A0}\;\alpha\;\beta-P_{A0}\;\alpha=P_{A0}(\beta-\alpha\beta-\alpha)$$
$$\Delta P=P_{A0}[(\beta(1-\alpha)-\alpha)]$$
Since $P_{A0}>0$, the term in between "[ ]" controls the sign of $\Delta P$, so we can evaluate 3 different outcomes.
Assuming just for a moment that $\Delta P=0$, we would get:
$$\beta(1-\alpha)-\alpha=0$$
Solving for $\beta$:
$$\beta=\frac{\alpha}{1-\alpha}=g(\alpha)$$
Then the three possible outcomes are:
(i) If $\beta>g(\alpha)$, then $\Delta P >0$
(ii) If $\beta=g(\alpha)$, then $\Delta P =0$
(iii) If $\beta<g(\alpha)$, then $\Delta P <0$
Now, we can compare $f(\alpha)$ vs $g(\alpha)$, noting that $f$ represents the actual value of $\beta$, and $g$ represents the exact boundary at which $\Delta P$ =0.
Considering $\alpha$ is a dissociation fraction, and the fact that for equilibrium to make sense neither 0% nor 100% of dissociation is possible, the domain of $\alpha$ is:
$$\alpha ∈ (0,1)$$
We can now tabulate $\alpha$ vs $f(\alpha)$ vs $g(\alpha)$:
\begin{array} {|r|r|}\hline \alpha & f(\alpha) & g(\alpha) \\ \hline \to 0 & \to 0 & \to 0 \\ \hline 0.1 & 0.06538 & 0.1111 \\ \hline 0.2 & 0.1386 & 0.25 \\ \hline 0.3 & 0.2206 & 0.4286 \\ \hline 0.4 & 0.3122 & 0.6667 \\ \hline 0.5 & 0.4142 & 1 \\ \hline 0.6 & 0.5262 & 1.5 \\ \hline 0.7 & 0.6463 & 2.333 \\ \hline 0.8 & 0.7703 & 4 \\ \hline 0.9 & 0.8912 & 9 \\ \hline \to 1 & \to 1 & \to \infty \\ \hline \end{array}
From the table, we can observe that for every value within the domain of $\alpha$, $f$ is smaller than $g$:
$$\alpha ∈ (0,1) \to f(\alpha)<g(\alpha)$$
In other words, condition (iii) is met, and $\Delta P$ is negative, so total equilibrium pressure decreases after removing all the $\ce{Cl2}$:
$$\beta<g(\alpha)\to \Delta P<0\to P_2<P_1$$
Although it's not very important, as a side note it also follows that:
$$\alpha ∈ (0,1) \to \beta<\alpha$$
I included a graph below to illustrate the solution of the problem, where I also added a $\beta=\alpha$ function as a reference line:
