Dirac being opaque and hard to follow? Well I never...
In Chapter 10 Dirac argues on physical grounds that the eigenkets of an observable must form a complete set. His argument goes that say we have an observable with eigenstates $|\varepsilon\rangle$ and some general state $|P\rangle$. Then I can write $|P\rangle$ as \begin{equation}|P\rangle = \sum a_\varepsilon|\varepsilon\rangle + \sum b_\gamma |\gamma\rangle \end{equation} Where $|\gamma\rangle$ are some state that cannot be written as a combination of the $|\varepsilon\rangle$s. Unlike Dirac I have normalized my eigenstates so that $\langle \varepsilon|\varepsilon\rangle = 1$ by pulling out a complex factor $a_\gamma$ and have not bothered to write down the integral, as the argument is essentially the same and it just complicates things. Now if we make a measurement of $\varepsilon$ for this state there is a probability $|b_\gamma|^2$ that we will find the system in each of the states $|\gamma\rangle$. But $|\gamma\rangle$ does not correspond to any allowed value of $\varepsilon$, so clearly does not make sense for the result of a measurement of $\varepsilon$! Therefore $b_\gamma = 0$ for all $\gamma$ and the $|\varepsilon\rangle$s form a complete set.
In the part you are asking about he is trying to prove that this way of writing $|P\rangle = \sum a_\varepsilon|\varepsilon\rangle$ is unique, i.e. for a given state $|P\rangle$ there is only one possible set of coefficients $a_\varepsilon$. Now we have from orthogonality that \begin{equation}\langle\varepsilon^\prime|\varepsilon\rangle = \left\{\begin{array}{lc} 1& \varepsilon = \varepsilon^\prime \\ 0 & \varepsilon\ne\varepsilon^\prime\end{array}\right.\end{equation} Let us say that there is another set of coefficients $a_\varepsilon^\prime$ such that $|P\rangle = \sum a_\varepsilon^\prime|\varepsilon\rangle$. Subtracting these two expressions for $|P\rangle$ we find \begin{equation} 0 = \sum (a_\varepsilon - a_\varepsilon^\prime)|\varepsilon\rangle\end{equation} Multiplying by $\langle\varepsilon^\prime|$ we find that all the terms go to 0 except \begin{equation} 0 = (a_{\varepsilon^\prime}-a_{\varepsilon^\prime}^\prime)\end{equation} So it terns out $a_{\varepsilon^\prime} = a_{\varepsilon^\prime}^\prime$ and since this must be true for each $\varepsilon$, the expression for $|P\rangle$ was unique.
With the integral included the discrete coefficients are replaced by functions in the integral $a(\varepsilon)$ and the normalisation is $\langle\varepsilon^\prime|\varepsilon\rangle = \delta(\varepsilon^\prime - \varepsilon)$ but otherwise the argument is unchanged.