6. $\frac{\sqrt{2}}{2}$.

Let $\angle C A P=\alpha$.
By the problem, $\angle B A P=\frac{\pi}{2}-\alpha$.
$$\begin{array}{l} \text { Given }|\overrightarrow{A P}|=2, \overrightarrow{A P} \cdot \overrightarrow{A C}=2, \overrightarrow{A P} \cdot \overrightarrow{A B}=1 \\ \Rightarrow|\overrightarrow{A C}|=\frac{1}{\cos \alpha},|\overrightarrow{A B}|=\frac{1}{2 \sin \alpha} \\ \Rightarrow|\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A P}|^{2} \\ =|\overrightarrow{A B}|^{2}+|\overrightarrow{A C}|^{2}+|\overrightarrow{A P}|^{2}+2 \overrightarrow{A B} \cdot \overrightarrow{A C}+ \\ 2 \overrightarrow{A B} \cdot \overrightarrow{A P}+2 \overrightarrow{A C} \cdot \overrightarrow{A P} \\ =\frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{4 \sin ^{2} \alpha}+\frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{\cos ^{2} \alpha}+10 \\ =\frac{\cos ^{2} \alpha}{4 \sin ^{2} \alpha}+\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}+\frac{45}{4} \\ \geqslant 2 \sqrt{\frac{\cos ^{2} \alpha}{4 \sin ^{2} \alpha} \cdot \frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}}+\frac{45}{4}=\frac{49}{4}, \end{array}$$

When and only when $\frac{\cos ^{2} \alpha}{4 \sin ^{2} \alpha}=\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}$, i.e., $\tan \alpha=\frac{\sqrt{2}}{2}$, $|\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A P}|$ achieves its minimum value $\frac{7}{2}$.